confusion when comparing char * with a string literal

[william]

below is a short piece of code I wrote to testify my understanding of
char *, and array.

#include <stdio.h>

int main()
{
char *str=NULL;
char x[]="today is good!";
printf("%s", str);
str=strtok(x," ");
if (str=="today") //<==here is line that confuses me
printf("they equals!\n");
return 0;
}

I printed "str" first, and the console displayed "today". However,
when I try to comapare 'str' with "today", the condition failed!

Exactly speaking, I know that 'str' is a 4 byte pointer of char type,
so it is not equal to a string. But even in the gdb, I used 'p str',
it printed "today".

So, my question is how do we compare arrays and char *(I found that it
is so commonly used as string)?

one more question: is it true that we have to initialize a char * or
points it to somewhere in memory, or to malloc memory to it before we
can use it?
Because I got segfault several times arising from this problem too.

Thank you for your reply in advance!

Ji

 

[william]

below is a short piece of code I wrote to testify my understanding of
char *, and array.

#include <stdio.h>

int main()
{
char *str=NULL;
char x[]="today is good!";
printf("%s", str);
str=strtok(x," ");
if (str=="today") //<==here is line that confuses me
printf("they equals!\n");
return 0;
}

I printed "str" first, and the console displayed "today". However,
when I try to comapare 'str' with "today", the condition failed!

Exactly speaking, I know that 'str' is a 4 byte pointer of char type,
so it is not equal to a string. But even in the gdb, I used 'p str',
it printed "today".

So, my question is how do we compare arrays and char *(I found that it
is so commonly used as string)?

one more question: is it true that we have to initialize a char * or
points it to somewhere in memory, or to malloc memory to it before we
can use it?
Because I got segfault several times arising from this problem too.

Thank you for your reply in advance!

Ji

By the way, gcc also reported:
warning: assignment makes pointer from integer without a cast
on the line: str=strtok(x," ");
where I think the type is compatible :-(

 

[santosh]

below is a short piece of code I wrote to testify my understanding of
char *, and array.

#include <stdio.h>

Include string.h to get the prototype for strtok.

int main()
{
char *str=NULL;
char x[]="today is good!";
printf("%s", str);

Huh? You're passing a null pointer to printf and telling it to follow
it? I think you meant to initilaise it to point to x.

str = x;

str=strtok(x," ");

strtok need a char * as it's first argument. x is a constant pointer.

if (str=="today") //<==here is line that confuses me

You cannot directly compare strings in C. You'll need to compare them
element by element. The Standard C library function for this is strcmp
or strncmp.

printf("they equals!\n");
return 0;
}

I printed "str" first, and the console displayed "today". However,
when I try to comapare 'str' with "today", the condition failed!

Exactly speaking, I know that 'str' is a 4 byte pointer of char type,

No. The Standard says nothing about the sizes of pointer types. Nor do
you need to assume a particular size for what you're doing. And it's
not a pointer *of* type char. It's a pointer *to* type char.

So, my question is how do we compare arrays and char *(I found that it
is so commonly used as string)?
strcmp/strncmp/memcmp

one more question: is it true that we have to initialize a char * or
points it to somewhere in memory, or to malloc memory to it before we
can use it?
Yes.

Because I got segfault several times arising from this problem too.

Indeed you'll if you work on pointers that don't point to any valid
object.

 

[Ben Pfaff]

char *str=NULL;
char x[]="today is good!";
printf("%s", str);

str=strtok(x," ");

strtok need a char * as it's first argument. x is a constant pointer.

You're off-base here. There's nothing wrong with passing x to
strtok as the first argument (modulo all the problems that are
endemic with strtok anyhow).

By the way: x is an array, not a constant pointer.

 

[Chris Dollin]

below is a short piece of code I wrote to testify my understanding of
char *, and array.

#include <stdio.h>

int main()
{
char *str=NULL;
char x[]="today is good!";
printf("%s", str);
str=strtok(x," ");
if (str=="today") //<==here is line that confuses me
printf("they equals!\n");
return 0;
}

I printed "str" first, and the console displayed "today". However,
when I try to comapare 'str' with "today", the condition failed!

Exactly speaking, I know that 'str' is a 4 byte pointer of char type,

The number of bytes is takes up isn't portable.

so it is not equal to a string. But even in the gdb, I used 'p str',
it printed "today".

gdb (which is off-topic) is being helpful.

So, my question is how do we compare arrays and char *(I found that it
is so commonly used as string)?
strcmp().

one more question: is it true that we have to initialize a char * or
points it to somewhere in memory, or to malloc memory to it before we
can use it?

Yes.

If it's declared as a pointer, it should point /somewhere/ (or be
null) to make sense using it.

Because I got segfault several times arising from this problem too.

You were lucky. (Unlucky would be it /not/ faulting, until what you
though was a working program fell over mysteriously while demonstrating
to an Important Customer in the presence of Godlike Manager.)

 

[Chris Dollin]

on the line: str=strtok(x," ");
where I think the type is compatible :-(

You didn't #include <string.h>, so there's no declaration for
`strtok`, so the compiler assumes return-type `int`, which you
can't convert to char* without a cast.

/Do not/ put in a cast. Put in the #include.

 

[Keith Thompson]

below is a short piece of code I wrote to testify my understanding of
char *, and array.

#include <stdio.h>

int main()
{
char *str=NULL;
char x[]="today is good!";
printf("%s", str);
str=strtok(x," ");
if (str=="today") //<==here is line that confuses me
printf("they equals!\n");
return 0;
}

I printed "str" first, and the console displayed "today". However,
when I try to comapare 'str' with "today", the condition failed!

Exactly speaking, I know that 'str' is a 4 byte pointer of char type,
so it is not equal to a string. But even in the gdb, I used 'p str',
it printed "today".

Apparently gdb didn't print the actual value of str (which would be an
address); it chose to display the string that it points to, which
presumably is more useful.

So, my question is how do we compare arrays and char *(I found that it
is so commonly used as string)?

one more question: is it true that we have to initialize a char * or
points it to somewhere in memory, or to malloc memory to it before we
can use it?
Because I got segfault several times arising from this problem too.

It depends on what you're using it for. If you're treating it as a
pointer to a string, it needs to point to a string, which means you
need to allocate memory to hold the string.

There's another problem in your program; I'll explain in another
followup.

 

[Keith Thompson]

below is a short piece of code I wrote to testify my understanding of
char *, and array.

#include <stdio.h>

int main()
{
char *str=NULL;
char x[]="today is good!";
printf("%s", str);
str=strtok(x," ");
if (str=="today") //<==here is line that confuses me
printf("they equals!\n");
return 0;
}

[...]

By the way, gcc also reported:
warning: assignment makes pointer from integer without a cast
on the line: str=strtok(x," ");
where I think the type is compatible :-(

You know that strtok() returns a result of type char*, but how is the
compiler supposed to know that? You haven't provided a declaration of
the strtok() function, so when the compiler sees a call to it, it
assumes (incorrectly, in this case), that it returns int. (That's in
C90; the rules are more strict in C99.)

The way to provide a declaration for strtok() is to add
#include <string.h>
to the top of your program.

Whenever you use a library function, or any declaration in the
library, you need to provide a #include directive for the header that
declares it.

 

[anilnf]

below is a short piece of code I wrote to testify my understanding of
char *, and array.

#include <stdio.h>

int main()
{
char *str=NULL;
char x[]="today is good!";
printf("%s", str);
str=strtok(x," ");
if (str=="today") //<==here is line that confuses me
printf("they equals!\n");
return 0;
}

I printed "str" first, and the console displayed "today". However,
when I try to comapare 'str' with "today", the condition failed!

Exactly speaking, I know that 'str' is a 4 byte pointer of char type,
so it is not equal to a string. But even in the gdb, I used 'p str',
it printed "today".

So, my question is how do we compare arrays and char *(I found that it
is so commonly used as string)?

one more question: is it true that we have to initialize a char * or
points it to somewhere in memory, or to malloc memory to it before we
can use it?
Because I got segfault several times arising from this problem too.

Thank you for your reply in advance!

Ji

By the way, gcc also reported:
warning: assignment makes pointer from integer without a cast
on the line: str=strtok(x," ");
where I think the type is compatible :-(

Why does it print "today" ? :

char * str means ..u know ..: " that 'str' is a 4 byte pointer
of char type"
But what ur program does is it makes a pointer variable, to which u
havn't given any memory address..so if u try to access the memory
location pointed to by ur pointer ,it will access some random
memory .. Here itz the place were u hav stored the array "today " ..If
it accesses some memory location outside the limits alloted for ur
program ,ur OS will hook it by saying 'segfault'.. Can't u use 'strcmp': in string.h ... U 'll need to compare
each character seperately to compare two arrays... Yes ... 'char * ' is just a pointer ..It has to point somewhere
before u could access content from it.. Special to gcc... In gcc u won't be able to change a char []
declared unless u use --enable options as argument to gcc ...Moreover
itz a warning leave it ...(I think so ..but to err is human)


-Nf

 

[Flash Gordon]

below is a short piece of code I wrote to testify my understanding of
char *, and array.
#include <stdio.h>
int main()
{
char *str=NULL;
char x[]="today is good!";
printf("%s", str);
str=strtok(x," ");
if (str=="today") //<==here is line that confuses me
printf("they equals!\n");
return 0;
}
I printed "str" first, and the console displayed "today". However,
when I try to comapare 'str' with "today", the condition failed!
Exactly speaking, I know that 'str' is a 4 byte pointer of char type,
so it is not equal to a string. But even in the gdb, I used 'p str',
it printed "today".
So, my question is how do we compare arrays and char *(I found that it
is so commonly used as string)?
one more question: is it true that we have to initialize a char * or
points it to somewhere in memory, or to malloc memory to it before we
can use it?
Because I got segfault several times arising from this problem too.
Thank you for your reply in advance!
Ji

By the way, gcc also reported:
warning: assignment makes pointer from integer without a cast
on the line: str=strtok(x," ");
where I think the type is compatible :-(

Why does it print "today" ? :

char * str means ..u know ..: " that 'str' is a 4 byte pointer
of char type"

No it does not. It means str is a pointer to char, there is nothing to
suggest how large the pointer is.

Also, please don't use contractions like u and ur, it makes reading your
post difficult, and more people read each post that write it.

But what ur program does is it makes a pointer variable, to which u
havn't given any memory address..so if u try to access the memory
location pointed to by ur pointer ,it will access some random
memory .. Here itz the place were u hav stored the array "today " ..If
it accesses some memory location outside the limits alloted for ur
program ,ur OS will hook it by saying 'segfault'..

Or it might not. It could do anything.

Can't u use 'strcmp': in string.h ... U 'll need to compare
each character seperately to compare two arrays...
Yes ... 'char * ' is just a pointer ..It has to point somewhere
before u could access content from it..
Special to gcc...

No it is not.

> In gcc u won't be able to change a char []

Complete rubbish. A char array is modifiable unless declared as const,
string literals are a completely different matter (you are not allowed
to modify them) but since the OP did NOT try to modify a string literal
they are not relevant.

declared unless u use --enable options as argument to gcc ...Moreover
itz a warning leave it ...(I think so ..but to err is human)

Incredibly stupid advice. The warning was due to a very serious error,
one whic *will* cause the program to fail on some modern C
implementations. Namely failing to provide a prototype for strtok, and
the best solution to this is including the standard header that provides it.

Ignoring warnings is NEVER a good idea. Very occasionally it makes sense
to leave in a warning, but only when you have enough experience to know
*why* you are write in that specific situation.

 

[Joe Wright]

char *str=NULL;
char x[]="today is good!";
printf("%s", str);
str=strtok(x," ");

strtok need a char * as it's first argument. x is a constant pointer.

You're off-base here. There's nothing wrong with passing x to
strtok as the first argument (modulo all the problems that are
endemic with strtok anyhow).

By the way: x is an array, not a constant pointer.

Once strtok gets it, x is a char* with no indication whether it started
out as an array of char or a pointer to constant string.

How tough is it living in Palo Alto? :=)

 

[Ben Pfaff]

How tough is it living in Palo Alto? :=)

Very, very expensive.

 

[Joe Wright]

Very, very expensive.

I left the area 15 years ago and it was expensive then. Now I live in
Northern Virginia within stone's throw of the Pentagon. How expensive is
it? A decent meal for two costs between 60 and 160 dollars (or more).

The average drink is 5 or 6 dollars in any old bar. Is it worse than
that in Silicon Valley?

 

[Richard Heathfield]

Joe Wright said:

Now I live in Northern Virginia within stone's throw of the Pentagon.

Nevertheless, I recommend that you resist the temptation.

 

[Joe Wright]

Joe Wright said:


Nevertheless, I recommend that you resist the temptation.

Not to worry. There are no stones within stone's throw of the Pentagon.
Security, I suppose.

 

[CBFalconer]

Not to worry. There are no stones within stone's throw of the Pentagon.
Security, I suppose.

I imagine getting caught smuggling contraband stones to within a
stones throw of the Pentagon will get you thrown into the Gitmo
gulag for an indefinite stay in stone cells, together with stony
inquisitions.

 

[william]

(e-mail address removed) wrote, On 17/03/07 19:29:

below is a short piece of code I wrote to testify my understanding of
char *, and array.
#include <stdio.h>
int main()
{
char *str=NULL;
char x[]="today is good!";
printf("%s", str);
str=strtok(x," ");
if (str=="today") //<==here is line that confuses me
printf("they equals!\n");
return 0;
}
I printed "str" first, and the console displayed "today". However,
when I try to comapare 'str' with "today", the condition failed!
Exactly speaking, I know that 'str' is a 4 byte pointer of char type,
so it is not equal to a string. But even in the gdb, I used 'p str',
it printed "today".
So, my question is how do we compare arrays and char *(I found that it
is so commonly used as string)?
one more question: is it true that we have to initialize a char * or
points it to somewhere in memory, or to malloc memory to it before we
can use it?
Because I got segfault several times arising from this problem too.
Thank you for your reply in advance!
Ji
By the way, gcc also reported:
warning: assignment makes pointer from integer without a cast
on the line: str=strtok(x," ");
where I think the type is compatible :-(
Why does it print "today" ? :

char * str means ..u know ..: " that 'str' is a 4 byte pointer
of char type"

No it does not. It means str is a pointer to char, there is nothing to
suggest how large the pointer is.

Also, please don't use contractions like u and ur, it makes reading your
post difficult, and more people read each post that write it.

But what ur program does is it makes a pointer variable, to which u
havn't given any memory address..so if u try to access the memory
location pointed to by ur pointer ,it will access some random
memory .. Here itz the place were u hav stored the array "today " ..If
it accesses some memory location outside the limits alloted for ur
program ,ur OS will hook it by saying 'segfault'..

Or it might not. It could do anything.

Can't u use 'strcmp': in string.h ... U 'll need to compare
each character seperately to compare two arrays...
Yes ... 'char * ' is just a pointer ..It has to point somewhere
before u could access content from it..
Special to gcc...

No it is not.

In gcc u won't be able to change a char []

Complete rubbish. A char array is modifiable unless declared as const,
string literals are a completely different matter (you are not allowed
to modify them) but since the OP did NOT try to modify a string literal
they are not relevant.

declared unless u use --enable options as argument to gcc ...Moreover
itz a warning leave it ...(I think so ..but to err is human)

Incredibly stupid advice. The warning was due to a very serious error,
one whic *will* cause the program to fail on some modern C
implementations. Namely failing to provide a prototype for strtok, and
the best solution to this is including the standard header that provides it.

Ignoring warnings is NEVER a good idea. Very occasionally it makes sense
to leave in a warning, but only when you have enough experience to know
*why* you are write in that specific situation.
--

Thank you for this piece of advice: "DO NOT ingore warnings"

 

[william]

Thank you, everyone speaking here. Thank you for your replies and
effort, which helped me to understand these portion of C much better.

and sorry for multi-post in both c/c++ groups.

 

[CBFalconer]

.... snip ...

Thank you for this piece of advice: "DO NOT ingore warnings"

No, I recommend exgoreing them. Alternatively, ask Al.