# Can anyone improve this any further......

Discussion in 'C Programming' started by Sean Kenwrick, Feb 6, 2004.

1. ### Sean KenwrickGuest

I am writing a byte-code interpreter/emulator for a language that
exclusively uses strings for variables (i.e all variables are pushed onto
the stack as strings). Therefore for arithmetic operations I need to
convert the string to a 32 bit integer, carry out the arithmetic operation,
then convert it back to a string before pushing it back onto the stack.
This can happen millions of times per second so it is essential that I have
optimised (for speed) atol() and ltoa() functions.

Here is my attempt at ltoa()... (The __fastcall keyword is to tell my
compiler to pass arguments through registers and is used instead of inline
because it seems to be just as fast and this function gets called from
various modules ..).

static char * __fastcall myltoa(long n,char * s)
{
register long r, k;
register int flag = 0;
register int next;

next = 0;
if (n == 0) {
s[next++] = '0';
}
else {
if (n < 0) {
s[next++] = '-';
n = -n;
}

if(n < 100) goto label2;
if(n < 100000) goto label1;

k = 1000000000;
r = n/k;
if(flag) s[next++] = '0' + r;
else if(r > 0){ s[next++] = '0' + r;flag = 1;}
n -= r * k;

k = 100000000;
r = n/k;
if(flag) s[next++] = '0' + r;
else if(r > 0){ s[next++] = '0' + r;flag = 1;}
n -= r * k;

k = 10000000;
r = n/k;
if(flag) s[next++] = '0' + r;
else if(r > 0){ s[next++] = '0' + r;flag = 1;}
n -= r * k;

k = 1000000;
r = n/k;
if(flag) s[next++] = '0' + r;
else if(r > 0){ s[next++] = '0' + r;flag = 1;}
n -= r * k;

k = 100000;
r = n/k;
if(flag) s[next++] = '0' + r;
else if(r > 0){ s[next++] = '0' + r;flag = 1;}
n -= r * k;
label1:
k = 10000;
r = n/k;
if(flag) s[next++] = '0' + r;
else if(r > 0){ s[next++] = '0' + r;flag = 1;}
n -= r * k;

k = 1000;
r = n/k;
if(flag) s[next++] = '0' + r;
else if(r > 0){ s[next++] = '0' + r;flag = 1;}
n -= r * k;

k = 100;
r = n/k;
if(flag) s[next++] = '0' + r;
else if(r > 0){ s[next++] = '0' + r;flag = 1;}
n -= r * k;
label2:
k = 10;
r = n/k;
if(flag) s[next++] = '0' + r;
else if(r > 0){ s[next++] = '0' + r;flag = 1;}
n -= r * k;

r=n;
s[next++] = '0' + r;
}
s[next] = '\0';
return(s);
}

The goto's are there because I recognised that the majority of numbers are
less than 100 (mainly return codes from functions) and those that are not
less than 100 tend to be less than 100,000 (this language is not usually
used for mathemetics). This did give me another 15-20% improvement in
speed on a tight loop going some arithmetic..

By the way, I have tried pre-calculating integer values for all strings
pushed onto the stack (by always appending the 4 byte integer value on the
stack at the end of the string, but it actually caused a slowdown since the
I have to do a atol() on every function return value or string concatenation
operation even if the value is not going to be used in an arithmetic or
comparison operation.

Any hints would be welcome...

Sean

Sean Kenwrick, Feb 6, 2004

2. ### peteGuest

Sean Kenwrick wrote:
>
> I am writing a byte-code interpreter/emulator for a language that
> exclusively uses strings for variables (i.e all variables are pushed onto
> the stack as strings). Therefore for arithmetic operations I need to
> convert the string to a 32 bit integer, carry out the arithmetic operation,
> then convert it back to a string before pushing it back onto the stack.
> This can happen millions of times per second so it is essential that I have
> optimised (for speed) atol() and ltoa() functions.
>
> Here is my attempt at ltoa()... (The __fastcall keyword is to tell my
> compiler to pass arguments through registers and is used instead of inline
> because it seems to be just as fast and this function gets called from
> various modules ..).
>
> static char * __fastcall myltoa(long n,char * s)
> {
> register long r, k;
> register int flag = 0;
> register int next;
>
> next = 0;
> if (n == 0) {
> s[next++] = '0';
> }
> else {
> if (n < 0) {
> s[next++] = '-';
> n = -n;

Undefined behavior if (-LONG_MAX > n)

> }
>
> if(n < 100) goto label2;
> if(n < 100000) goto label1;
>
> k = 1000000000;
> r = n/k;
> if(flag) s[next++] = '0' + r;

if(flag)
flag is already known to be zero, at this point.

> else if(r > 0){ s[next++] = '0' + r;flag = 1;}

> Any hints would be welcome...

#include <limits.h>
char *lto_a(long n, char *s)
{
char c, *p, *q;
int flag = 0;

q = p = s;
if (0 > n) {
*p++ = '-';
++s;
if (-LONG_MAX > n) {
flag = 1;
n = LONG_MAX;
} else {
n = -n;
}
}
do {
*p++ = (char)(n % 10 + '0');
n /= 10;
} while (n != 0);
if (flag) {
++*s;
}
for (*p = '\0'; --p > s; ++s) {
c = *s;
*s = *p;
*p = c;
}
return q;
}

--
pete

pete, Feb 6, 2004

3. ### Sean KenwrickGuest

"pete" <> wrote in message
news:...
> Sean Kenwrick wrote:
> >
> > I am writing a byte-code interpreter/emulator for a language that
> > exclusively uses strings for variables (i.e all variables are pushed

onto
> > the stack as strings). Therefore for arithmetic operations I need to
> > convert the string to a 32 bit integer, carry out the arithmetic

operation,
> > then convert it back to a string before pushing it back onto the stack.
> > This can happen millions of times per second so it is essential that I

have
> > optimised (for speed) atol() and ltoa() functions.
> >
> > Here is my attempt at ltoa()... (The __fastcall keyword is to tell my
> > compiler to pass arguments through registers and is used instead of

inline
> > because it seems to be just as fast and this function gets called from
> > various modules ..).
> >
> > static char * __fastcall myltoa(long n,char * s)
> > {
> > register long r, k;
> > register int flag = 0;
> > register int next;
> >
> > next = 0;
> > if (n == 0) {
> > s[next++] = '0';
> > }
> > else {
> > if (n < 0) {
> > s[next++] = '-';
> > n = -n;

>
> Undefined behavior if (-LONG_MAX > n)
>
> > }
> >
> > if(n < 100) goto label2;
> > if(n < 100000) goto label1;
> >
> > k = 1000000000;
> > r = n/k;
> > if(flag) s[next++] = '0' + r;

>
> if(flag)
> flag is already known to be zero, at this point.
>
> > else if(r > 0){ s[next++] = '0' + r;flag = 1;}

>
>
> > Any hints would be welcome...

>
> Please time this for me:
>
> #include <limits.h>
> char *lto_a(long n, char *s)
> {
> char c, *p, *q;
> int flag = 0;
>
> q = p = s;
> if (0 > n)

> *p++ = '-';
> ++s;
> if (-LONG_MAX > n) {
> flag = 1;
> n = LONG_MAX;
> } else {
> n = -n;
> }
> }
> do {
> *p++ = (char)(n % 10 + '0');
> n /= 10;
> } while (n != 0);
> if (flag) {
> ++*s;
> }
> for (*p = '\0'; --p > s; ++s) {
> c = *s;
> *s = *p;
> *p = c;
> }
> return q;
> }
>
> --
> pete

After I change the call to __fastcall and use register variables It's very
slightly slower (mine took 4650ms, yours took 4860ms for a 1,000,000 times
loop doing some simple arithmetic). The difference is probably in the
strrev() required at the end - but I like the solution of going from the
right to left and I think I have a version that might not need the strrev()
at the end..... I will post my attempt and the timing later...

Thanks
Sean

Sean Kenwrick, Feb 6, 2004
4. ### Francois GrieuGuest

#include <limits.h>

#if INT_MAX==0x7FFFFFFF && INT_MIN+2147483647>=-1
/* 32 bit signed arithmetic assumed */
char *myltoa(long n, char *s)
{
unsigned short j = 1;
if (n<0)
{ /* handle negatives */
s[0] = '-'; ++j;
if (INT_MIN+2147483647==-1 && n==INT_MIN)
{ /* handle -2147483648 the hard way */
s[1] = '2'; ++j;
n = -147483648;
}
n = -n;
}
/* find length using kinda dichotomy */
s[j += (n<10000)?(n<100)?(n>=10)n>=1000)+2:
(n<100000000)?(n<1000000)?(n>=100000)+4:
(n>=10000000)+6n>=1000000000)+8] = '\0';
/* convert from right to left */
do s[--j] = n%10+'0'; while ((n/=10)!=0);
/* all done */
return(s);
}
#endif

François Grieu

Francois Grieu, Feb 6, 2004
5. ### peteGuest

Sean Kenwrick wrote:

> After I change the call to __fastcall and use register variables
> It's very slightly slower (mine took 4650ms,
> yours took 4860ms for a 1,000,000 times
> loop doing some simple arithmetic).
> The difference is probably in the strrev() required at the end
> - but I like the solution of going from the
> right to left and I think I have a version that might not need
> the strrev() at the end.....
> I will post my attempt and the timing later...

Don't forget the (-LONG_MAX > n) case.
Thank you.

--
pete

pete, Feb 6, 2004
6. ### Sean KenwrickGuest

"Sean Kenwrick" <> wrote in message
news:c00ing\$4dc\$...
>
> "pete" <> wrote in message
> news:...
> > Sean Kenwrick wrote:
> > >
> > > I am writing a byte-code interpreter/emulator for a language that
> > > exclusively uses strings for variables (i.e all variables are pushed

> onto
> > > the stack as strings). Therefore for arithmetic operations I need

to
> > > convert the string to a 32 bit integer, carry out the arithmetic

> operation,
> > > then convert it back to a string before pushing it back onto the

stack.
> > > This can happen millions of times per second so it is essential that I

> have
> > > optimised (for speed) atol() and ltoa() functions.
> > >
> > > Here is my attempt at ltoa()... (The __fastcall keyword is to tell my
> > > compiler to pass arguments through registers and is used instead of

> inline
> > > because it seems to be just as fast and this function gets called from
> > > various modules ..).
> > >
> > > static char * __fastcall myltoa(long n,char * s)
> > > {
> > > register long r, k;
> > > register int flag = 0;
> > > register int next;
> > >
> > > next = 0;
> > > if (n == 0) {
> > > s[next++] = '0';
> > > }
> > > else {
> > > if (n < 0) {
> > > s[next++] = '-';
> > > n = -n;

> >
> > Undefined behavior if (-LONG_MAX > n)
> >
> > > }
> > >
> > > if(n < 100) goto label2;
> > > if(n < 100000) goto label1;
> > >
> > > k = 1000000000;
> > > r = n/k;
> > > if(flag) s[next++] = '0' + r;

> >
> > if(flag)
> > flag is already known to be zero, at this point.
> >
> > > else if(r > 0){ s[next++] = '0' + r;flag = 1;}

> >
> >
> > > Any hints would be welcome...

> >
> > Please time this for me:
> >
> > #include <limits.h>
> > char *lto_a(long n, char *s)
> > {
> > char c, *p, *q;
> > int flag = 0;
> >
> > q = p = s;
> > if (0 > n)

>
> > *p++ = '-';
> > ++s;
> > if (-LONG_MAX > n) {
> > flag = 1;
> > n = LONG_MAX;
> > } else {
> > n = -n;
> > }
> > }
> > do {
> > *p++ = (char)(n % 10 + '0');
> > n /= 10;
> > } while (n != 0);
> > if (flag) {
> > ++*s;
> > }
> > for (*p = '\0'; --p > s; ++s) {
> > c = *s;
> > *s = *p;
> > *p = c;
> > }
> > return q;
> > }
> >
> > --
> > pete

>
> After I change the call to __fastcall and use register variables It's very
> slightly slower (mine took 4650ms, yours took 4860ms for a 1,000,000 times
> loop doing some simple arithmetic). The difference is probably in the
> strrev() required at the end - but I like the solution of going from the
> right to left and I think I have a version that might not need the

strrev()
> at the end..... I will post my attempt and the timing later...
>
> Thanks
> Sean
>
>

By the way I mis-remembered the timings above, they were actually 2650 and
2860ms...

Here is the version I ended up with of the above algorithm. I can be
certain that the long passed does not exceed +/- LONG_MAX so I don't need
these tests, and I've done away with the strrev() at the end by filling the
string backwards from the 11th char and then returning a pointer to where
the actual number starts in the string..

char * __fastcall myltoa(long n, char *s)
{
register char *p;
register int flag=0;

p=s+12;
*p--='\0'; /* Null terminate */

// If n is negative
if (0 > n) {
flag=1;
n = -n;
}

do {
// Next char is n%10 (Rightmost digit)
*p-- = (char)(n % 10 + '0');
// Strip off the rightmost digit
n /= 10;
} while (n != 0); // Until n is 0

if(flag)
*p-- = '-';
// Return the offset into the string where the number starts
return p+1;
}

However there was no significant change in the timings so I suspect that the
problem with the above is that it is doing two divides per digit (in fact
one modulo and 1 divide) - and divides are typically alot slower than all
the other arithmetic operations.. In my version I do 1 divide, 1 multiply
and 1 subtraction which might be a saving of quite a few cycles per digit...

I can't see any way of optimising this version any firther unless there is a
way to get rid of one of the divides...
Sean

Sean Kenwrick, Feb 6, 2004
7. ### Sean KenwrickGuest

"Francois Grieu" <> wrote in message
news:...
>
> #include <limits.h>
>
> #if INT_MAX==0x7FFFFFFF && INT_MIN+2147483647>=-1
> /* 32 bit signed arithmetic assumed */
> char *myltoa(long n, char *s)
> {
> unsigned short j = 1;
> if (n<0)
> { /* handle negatives */
> s[0] = '-'; ++j;
> if (INT_MIN+2147483647==-1 && n==INT_MIN)
> { /* handle -2147483648 the hard way */
> s[1] = '2'; ++j;
> n = -147483648;
> }
> n = -n;
> }
> /* find length using kinda dichotomy */
> s[j += (n<10000)?(n<100)?(n>=10)n>=1000)+2:
> (n<100000000)?(n<1000000)?(n>=100000)+4:
> (n>=10000000)+6n>=1000000000)+8] = '\0';
> /* convert from right to left */
> do s[--j] = n%10+'0'; while ((n/=10)!=0);
> /* all done */
> return(s);
> }
> #endif
>
>
> François Grieu

This is similar to my solution expect that I didn't bother checking against
INT_MIN/MAX (because I know that these can't be exceeded) and I didn't
bother filling the string from the first character (so no need for the
length calculation (I just return a pointer to the offset in the string).
But this suffers from the same problem that I had with my version (See
previous post) in that it effectively does two divides per digit and this
seems to be causing the function to be slower than my orignal by about 10%..

If there is any way to get rid of one of the divides then this would
probably be faster..

Sean

Sean Kenwrick, Feb 6, 2004
8. ### peteGuest

Sean Kenwrick wrote:

> However there was no significant change in the timings
> so I suspect that the problem with the above is that it
> is doing two divides per digit
> (in fact one modulo and 1 divide)
> - and divides are typically alot slower than all
> the other arithmetic operations.. In my version I do 1 divide,
> 1 multiply and 1 subtraction which might be a saving of quite
> a few cycles per digit...
>
> I can't see any way of optimising this version any
> firther unless there is a way to get rid of one of the divides...

There is a way, but is it faster ?

#include <stdlib.h>
ldiv_t ldiv(long numer, long denom);

--
pete

pete, Feb 6, 2004
9. ### Sean KenwrickGuest

"pete" <> wrote in message
news:...
> Sean Kenwrick wrote:
>
> > However there was no significant change in the timings
> > so I suspect that the problem with the above is that it
> > is doing two divides per digit
> > (in fact one modulo and 1 divide)
> > - and divides are typically alot slower than all
> > the other arithmetic operations.. In my version I do 1 divide,
> > 1 multiply and 1 subtraction which might be a saving of quite
> > a few cycles per digit...
> >
> > I can't see any way of optimising this version any
> > firther unless there is a way to get rid of one of the divides...

>
> There is a way, but is it faster ?
>
> #include <stdlib.h>
> ldiv_t ldiv(long numer, long denom);
>
> --
> pete

I got rid of the extra divide and replaced it with a multiply and subtract
(to do the modulo bit). It is now as fast as my original, and even
slightly faster in some cases since I don't need to short-cut numbers less
than 100,000 or 100 for example... It is also a much cleaner solution so I
will use this instead. Here is my final version...

char * __fastcall myltoa(long n, char *s)
{
register int flag=0;
register int x;

s+=12;
*s--='\0'; /* Null terminate */

/* If n is negative */
if (0 > n) {
flag=1;
n = -n;
}

do {
/* Next char is n%10 (Rightmost digit) */
x=n/10;
*s-- = n -(x*10) + '0';
/* Strip off the rightmost digit */
n = x;
} while (n != 0); /* Until n is 0 */

if(flag)
*s-- = '-';
/* Return the offset into the string where the number starts */
return s+1;
}

Sean

Sean Kenwrick, Feb 6, 2004
10. ### Francois GrieuGuest

In article <c00tva\$s3c\$>,
"Sean Kenwrick" <> wrote:

> I got rid of the extra divide and replaced it with a multiply and subtract
> (to do the modulo bit). It is now (faster)

In my experience, is unusual that % is slower than / is for unsigned
quatities. I guess your system has a problem with modulo of signed
quantities.
Since apparently you have no requirement that the string returned
starts where s does, maybe try:

/* 32 bit signed arithmetic assumed */
/* does not handle -2147483648 */
char *myltoa(long n, char *s)
{
char neg;
*(s += 11) = neg = '\0';
if (n<0)
{
n = -n; /* does not handle -2147483648 */
neg = '-';
}
do *--s = (unsigned)n%10+'0';
while ((n = (unsigned)n/10)!=0);
if (neg!='\0')
*--s = neg;
return(s);
}

François Grieu

Francois Grieu, Feb 7, 2004
11. ### Sean KenwrickGuest

"Francois Grieu" <> wrote in message
news:...
> In article <c00tva\$s3c\$>,
> "Sean Kenwrick" <> wrote:
>
> > I got rid of the extra divide and replaced it with a multiply and

subtract
> > (to do the modulo bit). It is now (faster)

>
> In my experience, is unusual that % is slower than / is for unsigned
> quatities. I guess your system has a problem with modulo of signed
> quantities.
> Since apparently you have no requirement that the string returned
> starts where s does, maybe try:
>
> /* 32 bit signed arithmetic assumed */
> /* does not handle -2147483648 */
> char *myltoa(long n, char *s)
> {
> char neg;
> *(s += 11) = neg = '\0';
> if (n<0)
> {
> n = -n; /* does not handle -2147483648 */
> neg = '-';
> }
> do *--s = (unsigned)n%10+'0';
> while ((n = (unsigned)n/10)!=0);
> if (neg!='\0')
> *--s = neg;
> return(s);
> }
>
>
> François Grieu

The problem is that divide is typically ALOT slower than multiply or any
other arithmetic operations and this is probably the case for all CPUs.
And since the modulo operator requires a divide (I've checked the assembler
code) then it is just as slow as a normal divide. Your algorithm above
therefore still effectively does 2 divides per digit which is about 10%
slower than my very similar version of your algorithm that I wrote (see
previous post) which uses a multiple and a subtract instead of a modulo...

Sean

Sean Kenwrick, Feb 7, 2004
12. ### Johan LindhGuest

Sean Kenwrick wrote:

> "Francois Grieu" <> wrote in message
> news:...
>
>>In article <c00tva\$s3c\$>,
>> "Sean Kenwrick" <> wrote:
>>
>>
>>>I got rid of the extra divide and replaced it with a multiply and

>
> subtract
>
>>>(to do the modulo bit). It is now (faster)

>>
>>In my experience, is unusual that % is slower than / is for unsigned
>>quatities. I guess your system has a problem with modulo of signed
>>quantities.
>>Since apparently you have no requirement that the string returned
>>starts where s does, maybe try:
>>
>>/* 32 bit signed arithmetic assumed */
>>/* does not handle -2147483648 */
>>char *myltoa(long n, char *s)
>> {
>> char neg;
>> *(s += 11) = neg = '\0';
>> if (n<0)
>> {
>> n = -n; /* does not handle -2147483648 */
>> neg = '-';
>> }
>> do *--s = (unsigned)n%10+'0';
>> while ((n = (unsigned)n/10)!=0);
>> if (neg!='\0')
>> *--s = neg;
>> return(s);
>> }
>>
>>
>> François Grieu

>
>
> The problem is that divide is typically ALOT slower than multiply or any
> other arithmetic operations and this is probably the case for all CPUs.
> And since the modulo operator requires a divide (I've checked the assembler
> code) then it is just as slow as a normal divide. Your algorithm above
> therefore still effectively does 2 divides per digit which is about 10%
> slower than my very similar version of your algorithm that I wrote (see
> previous post) which uses a multiple and a subtract instead of a modulo...
>
> Sean
>
>

static long mask_list[] = { 1000000000, 100000000, 10000000, 1000000,
100000, 10000, 1000, 100, 10, 1 };
char * __fastcall myltoa2(long n, char *s)
{
char *org_s = s;
char c;

if( n <= 0 )
{
if( !n )
{
*s++ = '0';
*s = '\0';
return org_s;
}
n = -n;
*s++ = '-';
}

do
{
c = '0';
{
c++;
}
*s++ = c;
} while( n );

*s = '\0';

return org_s;
}

/Johan

Johan Lindh, Feb 7, 2004
13. ### peteGuest

Sean Kenwrick wrote:

> Here is my final version...
>
> char * __fastcall myltoa(long n, char *s)

Function identifiers starting with two underscores like that,
are reserved for the implementation.
Are you implementing C ?

N869
7.1.3 Reserved identifiers
[#1]

-- All identifiers that begin with an underscore and
either an uppercase letter or another underscore are
always reserved for any use.

--
pete

pete, Feb 7, 2004
14. ### Old WolfGuest

> > char * __fastcall myltoa(long n, char *s)
>
> Function identifiers starting with two underscores like that,
> are reserved for the implementation.
> Are you implementing C ?

No, but his compiler is implementing C-with-extensions,
one of which is the keyword "__fastcall".

Old Wolf, Feb 8, 2004
15. ### Sean KenwrickGuest

"Johan Lindh" <> wrote in message
news:...
> Sean Kenwrick wrote:
>
> > "Francois Grieu" <> wrote in message
> > news:...
> >
> >>In article <c00tva\$s3c\$>,
> >> "Sean Kenwrick" <> wrote:
> >>
> >>
> >>>I got rid of the extra divide and replaced it with a multiply and

> >
> > subtract
> >
> >>>(to do the modulo bit). It is now (faster)
> >>
> >>In my experience, is unusual that % is slower than / is for unsigned
> >>quatities. I guess your system has a problem with modulo of signed
> >>quantities.
> >>Since apparently you have no requirement that the string returned
> >>starts where s does, maybe try:
> >>
> >>/* 32 bit signed arithmetic assumed */
> >>/* does not handle -2147483648 */
> >>char *myltoa(long n, char *s)
> >> {
> >> char neg;
> >> *(s += 11) = neg = '\0';
> >> if (n<0)
> >> {
> >> n = -n; /* does not handle -2147483648 */
> >> neg = '-';
> >> }
> >> do *--s = (unsigned)n%10+'0';
> >> while ((n = (unsigned)n/10)!=0);
> >> if (neg!='\0')
> >> *--s = neg;
> >> return(s);
> >> }
> >>
> >>
> >> François Grieu

> >
> >
> > The problem is that divide is typically ALOT slower than multiply or any
> > other arithmetic operations and this is probably the case for all CPUs.
> > And since the modulo operator requires a divide (I've checked the

assembler
> > code) then it is just as slow as a normal divide. Your algorithm

above
> > therefore still effectively does 2 divides per digit which is about 10%
> > slower than my very similar version of your algorithm that I wrote (see
> > previous post) which uses a multiple and a subtract instead of a

modulo...
> >
> > Sean
> >
> >

>
> static long mask_list[] = { 1000000000, 100000000, 10000000, 1000000,
> 100000, 10000, 1000, 100, 10, 1 };
> char * __fastcall myltoa2(long n, char *s)
> {
> char *org_s = s;
> char c;
>
> if( n <= 0 )
> {
> if( !n )
> {
> *s++ = '0';
> *s = '\0';
> return org_s;
> }
> n = -n;
> *s++ = '-';
> }
>
>
> do
> {
> c = '0';
> while( n >= mask )
> {
> c++;
> }
> *s++ = c;
> } while( n );
>
> *s = '\0';
>
> return org_s;
> }
>
>
> /Johan
>
>

Your algorithm looked promising (faster) at first until I realised that I
was using numbers containing low-digit combinations in my test code (E.g.
a=1021020/324) . As soon as I used numbers with higher digit combinations
like 83894758 then the algorithm suffered with performance since it would
require 9 subtractions for the digit '9' etc. There is also a bug in the
algorithm as it was presented above. It need to take care of the case
where the number ends with trailing zeros (E.g 234000), so another loop
needed to be placed at the end as follows:

*s++='0';

Although the performance on average is quite similar to the current best
algorithm, I think I will stick to the version which has consistent
performance for any number and does not rely on digits being of low value
which could cause unexpected slowdowns...

Sean

Sean Kenwrick, Feb 9, 2004
16. ### Francois GrieuGuest

I now get the "* faster than %" point.
Can you try this, which I hope more pipeline-friendly?

#include <limits.h>
#if LONG_MAX==0x7FFFFFFFu && ULONG_MAX==0xFFFFFFFFu
/* 32 bit signed arithmetic assumed */
/* may not handle -2147483648, but will on most modern machines */
char *myltoa(long n, char *s)
{
unsigned long m;
*(s += 11) = '\0';
if (n>=0)
{
do
{
*--s = '0'+n-(m = (unsigned long)n/10u)*10u;
if (m==0)
break;
*--s = '0'+m-(unsigned long)(n = m/10u)*10u;
}
while (n!=0);
return s;
}
n = -n; /* may not handle -2147483648 */
do
{
*--s = '0'+n-(m = (unsigned long)n/10u)*10u;
if (m==0)
break;
*--s = '0'+m-(unsigned long)(n = m/10u)*10u;
}
while (n!=0);
*--s = '-';
return(s);
}
#endif

Next try will use floating point arithmetic.

François Grieu

Francois Grieu, Feb 10, 2004
17. ### Sean KenwrickGuest

"Francois Grieu" <> wrote in message
news:...
> I now get the "* faster than %" point.
> Can you try this, which I hope more pipeline-friendly?
>
> #include <limits.h>
> #if LONG_MAX==0x7FFFFFFFu && ULONG_MAX==0xFFFFFFFFu
> /* 32 bit signed arithmetic assumed */
> /* may not handle -2147483648, but will on most modern machines */
> char *myltoa(long n, char *s)
> {
> unsigned long m;
> *(s += 11) = '\0';
> if (n>=0)
> {
> do
> {
> *--s = '0'+n-(m = (unsigned long)n/10u)*10u;
> if (m==0)
> break;
> *--s = '0'+m-(unsigned long)(n = m/10u)*10u;
> }
> while (n!=0);
> return s;
> }
> n = -n; /* may not handle -2147483648 */
> do
> {
> *--s = '0'+n-(m = (unsigned long)n/10u)*10u;
> if (m==0)
> break;
> *--s = '0'+m-(unsigned long)(n = m/10u)*10u;
> }
> while (n!=0);
> *--s = '-';
> return(s);
> }
> #endif
>

For some reason the casts to unsigned caused this to slow down by 6% over
the current fastest. After removing the casts which appeared to be
unnecessary then it matched the time of my previous best exactly. Thus I
don't think the attempts to take advantage of pipelining had any effect and
the code thus became equivalent to my version (but slightly less

>
> Next try will use floating point arithmetic.
>
> François Grieu

I would be interested to see a solution based on FP arithmetic - perhaps FP
divides are quicker than the CPU integer divides?? I would be surprised
though...

Sean

Sean Kenwrick, Feb 10, 2004
18. ### Chris TorekGuest

[code using integer division etc]

In article <news:c0ai4a\$5fh\$>
Sean Kenwrick <> writes:
>For some reason the casts to unsigned caused this to slow down by 6% over
>the current fastest.

Some machines only have a "native" signed integer division, so
unsigned integer division requires "undoing" some work the
signed-divide instruction performed.

On other machines, there is no difference, or unsigned integer
division is slightly faster (e.g., pre-V8 SPARC, where there is
no divide instruction at all, and the .udiv routine can skip
the sign-fiddling).

>After removing the casts which appeared to be
>unnecessary then it matched the time of my previous best exactly.

The "first" cast may be necessary, or at least useful, depending
on the machine. If we ignore bizarre (but apparently legal)
implementations in which UINT_MAX is strictly less than INT_MAX or
-INT_MIN or both (where -INT_MIN means the mathematical value, not
the "C value"), we still have the very common case of two's complement
machines, in which -INT_MAX is off by one from INT_MIN: e.g.,
INT_MAX is 32767 while INT_MIN is (numerically) -32768, or INT_MAX
is 2147483647 while INT_MIN is -2147483648.

(Aside: if, in this example, INT_MIN is -32768, it has to be
expressed in C in some other form, such as (-32767 - 1), because
the integral constant expression -32768 consists of the unary "-"
followed by the constant 32768. But we just said INT_MAX is
32767 -- so the integral constant 32768 has type "unsigned int",
and negating it tells the compiler to compute the value mod 65536,
which happens to continue to be (unsigned int)32768. So -32768
and 32768U are the same number, on this machine. For 32-bit int
two's complement machines, one must write (-2147483647 - 1) or
similar, due to the same problem. The problem repeats for "long"
as well -- I have chosen to address only "int" here, even though
I think the original code uses "long".)

In any case, getting back to the point at hand, in C89 integer
division is allowed to round either towards 0 or towards -infinity,
so that (-3)/2 is either -1 (round towards 0) or -2 (round towards
-inf). The only constraint here is that ((a / b) * b) + (a % b)
should produce the original value in "a" (assuming b nonzero),
which in turn means that if (-3)/2 is -1, (-3)%2 must be -1 as
well, while if (-3)/2 is -2, (-3)%2 must be 1. (In C99, implementations
must round towards zero, I believe.)

Suppose, then, that INT_MIN is (-32767-1) and INT_MAX is 32767 and
the implementation rounds towards zero. Then suppose i is -32768
initially, and "i = -i" leaves it set to -32768 (which is quite
common even if it is undefined):

int i, j, rest;
...
rest = i / 10;
j = i % 10; /* or: j = i - (rest * 10); */

Since i is still -32768, this sets rest to -3276 (rounded towards
0) and j is -8. j cannot be converted to a digit by adding 0.

On the other hand, suppose the implementation rounds towards -inf.
Then rest is set to -3277 and j is 2, and converting j to a digit
gets '2'. If we do this in a loop, the next digit produced is '3'
(with rest set to -328), then '2' (-33), then '7' (-4), then '6';
and the number we will print is "-67232"!

By using unsigned arithmetic for the first division, we guarantee

int i, j, rest;
...
rest = -(unsigned)i / 10U;
j = (unsigned)i - (rest * 10U);

Now we divide 32768U by 10U giving 3276U, and then subtract 32760U
from 32768U to set j to 8. Converting to a digit gives '8' and
rest is now a positive integer well out of the problematic range.
The rest of the arithmetic can be done using signed divides, if
those are faster.

Since (I think) C99 guarantees rounding towards zero, the other
technique that can be used is to leave the negative number negative,
and simply adjust for the fact that j will be negative:

int i, j, rest;
bool negative = false; /* remember to #include <stdbool.h> */
char *p;
char buf[X]; /* use the log-base-8 trick to derive X */
...
p = buf + X;
*--p = '\0';
if (i < 0) {
negative = true;
rest = i / 10;
j = i - (rest * 10);

Now "rest" is at most INT_MIN/10 and j is (e.g.) -8 as before, so:

*--p = '0' + (-j);
i = -rest;
}
do {
rest = i / 10;
j = i - rest * 10;
*--p = '0' + j;
i = rest;
} while (i != 0);
if (negative)
*--p = '-';
printf("the result of conversion is %s\n", p);

This does assume that INT_MIN is less than -9, but the C standards
guarantee it.

>Thus I don't think the attempts to take advantage of pipelining
>[by swapping variables inside a doubled-up loop] had any effect and
>the code thus became equivalent to my version (but slightly less

If this kind of register-renaming pipelining will help, a good
compiler should expand the loop for you automatically anyway.

>I would be interested to see a solution based on FP arithmetic - perhaps FP
>divides are quicker than the CPU integer divides?? I would be surprised
>though...

Division is fundamentally more difficult than multiplication (see
comp.arch for gory hardware details), and for various reasons, many
hardware designers are much more willing to "spend" the transisitors
required to make hardware FP divide fast, than they are to spend
the transistors to make integer divide fast. So it can be the case
that FP divide happens in (say) 10 clocks while integer divide
takes (say) 100. But there are other tricks to compensate (such
as "reciprocal multiplication", providing the division is by a
constant), and again a good compiler really should do them for you.
The short version of the answer is "the speed of the operation is
off-topic in comp.lang.c." The long version starts with at
least this paragraph, and goes on for many more to discuss the
details of the CPU(s) and/or compiler(s) in question.
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) +1 801 277 2603
Reading email is like searching for food in the garbage, thanks to spammers.

Chris Torek, Feb 11, 2004
19. ### Dave ThompsonGuest

On Fri, 6 Feb 2004 12:24:01 +0000 (UTC), "Sean Kenwrick"
<> wrote:

> I am writing a byte-code interpreter/emulator for a language that
> exclusively uses strings for variables (i.e all variables are pushed onto
> the stack as strings). Therefore for arithmetic operations I need to
> convert the string to a 32 bit integer, carry out the arithmetic operation,
> then convert it back to a string before pushing it back onto the stack.
> This can happen millions of times per second so it is essential that I have
> optimised (for speed) atol() and ltoa() functions.

<snip>
> By the way, I have tried pre-calculating integer values for all strings
> pushed onto the stack (by always appending the 4 byte integer value on the
> stack at the end of the string, but it actually caused a slowdown since the
> I have to do a atol() on every function return value or string concatenation
> operation even if the value is not going to be used in an arithmetic or
> comparison operation.
>

If you can store numeric value as well as string value in variables --
assuming your language has variables (not pure functional or FORTHy)
-- how about pushing/storing etc. the numeric value *if known* and a
dummy value (or flag) if not; operations which need numeric value(s)
do the atol(s) if/when they encounter the dummy; similarly push/store
a dummy string, such as NULL, if only the numeric value is known, and
operations needing a string value do the ltoa if/when needed.

If you really do need to do ltoa an awful lot, your absolute best
performance is more likely achievable only by writing in assembler.
Especially since even if you write in C, the microoptimization results
often vary for different architectures, and even different models --
especially for x86, where division and different ways of
multiplication have bounced up and down through its history -- and
thus while you may have portable code, it isn't actually a portable

For x86, AIR the last time this was discussed on comp.lang.asm.x86,
the consensus for (all?) current CPUs was to use fstbp or else split
into 4-digit (< 16 - epsilon bit) chunks and 16:32 multiply by 64K/10
to form each digit (within each chunk). Or possibly for some models by
64K/100 and AAM to form digit pairs.

- David.Thompson1 at worldnet.att.net

Dave Thompson, Feb 14, 2004