Copy 4 byte block into a char array

[bg_ie]

Hi,

I have an array as follows -

char arr[100];

Now I wish to copy the following int -

int tmp = 0x01020304;

into this array, at element[12].

This following would have the same result -

ThePacket3[12] = 1;
ThePacket3[13] = 2;
ThePacket3[14] = 3;
ThePacket3[15] = 4;

How can I do this without breaking up my int into 4 bytes and assigning
each element one by one?

Thanks,

Barry.

 

[Tom St Denis]

ThePacket3[12] = 1;
ThePacket3[13] = 2;
ThePacket3[14] = 3;
ThePacket3[15] = 4;

How can I do this without breaking up my int into 4 bytes and assigning
each element one by one?

Write a function or macro.

It's called "modular design" where instead of copy/pasting the same
code snippet you found on snippets.org you actually build up a program
from a design where you abstract out common functionality to a body,
almost a "library" if you will, of support code for your application or
project.

Tom

 

[Nelu]

Hi,

I have an array as follows -

char arr[100];

Now I wish to copy the following int -

int tmp = 0x01020304;

into this array, at element[12].

This following would have the same result -

ThePacket3[12] = 1;
ThePacket3[13] = 2;
ThePacket3[14] = 3;
ThePacket3[15] = 4;

How can I do this without breaking up my int into 4 bytes and assigning
each element one by one?

You could probably try this:

memcpy(&arr[12],&tmp,sizeof(int));

Just make sure you know exactly what you're doing. For example int
doesn't need to be 4 byte long.

 

[Stephen Sprunk]

I have an array as follows -

char arr[100];

Now I wish to copy the following int -

int tmp = 0x01020304;

into this array, at element[12].

This following would have the same result -

ThePacket3[12] = 1;
ThePacket3[13] = 2;
ThePacket3[14] = 3;
ThePacket3[15] = 4;

How can I do this without breaking up my int into 4 bytes and assigning
each element one by one?

You can't, portably.

1. The endianness of a particular machine may cause the bytes to be assigned
to different chars than what you've listed above.
2. element[12] may not have proper alignment for an int, causing a crash.
3. sizeof(int) may not be 4, causing more than 4 elements of your array to
be modified.

This is why folks write macros (which assign the bytes in a specific order)
for such needs.

S

 

[James]

int tmp = 0x01020304;

into this array, at element[12].

This following would have the same result -

ThePacket3[12] = 1;
ThePacket3[13] = 2;
ThePacket3[14] = 3;
ThePacket3[15] = 4;

How can I do this without breaking up my int into 4 bytes and assigning
each element one by one?

if you know that you're on a big-endian machine and sizeof char * 4 ==
sizeof int, you could theoretically do:

int *slice;
slice = &arr[12];
*slice = tmp;

but that's obviously unportable as hell

 

[Tom St Denis]

You can't, portably.

....sorta.

In my projects I try to autodetect certain platforms and use memcpy if
possible, otherwise yeah I extract bytes and store in order.

Tom

 

[Keith Thompson]

I have an array as follows -

char arr[100];

Now I wish to copy the following int -

int tmp = 0x01020304;

into this array, at element[12].

This following would have the same result -

ThePacket3[12] = 1;
ThePacket3[13] = 2;
ThePacket3[14] = 3;
ThePacket3[15] = 4;

How can I do this without breaking up my int into 4 bytes and assigning
each element one by one?

int isn't necessarily 4 bytes, bytes aren't necessarily 8 bits, and
the value 0x01020304 isn't necessarily stored in the order 1, 2, 3, 4;
it's commonly 4, 3, 2, 1, but it could theoretically be in any of the
24 possible orders.

If you want to break the value of tmp down into 4 8-bit quantities,
and store then in successive elements of your array (it it called
"arr" or "ThePacket3"?), the only portable way to do it is by copying
each byte individually, after extracting them using shifts and masks.

If you're willing to make non-portable assumptions about how an int is
represented, you can use memcpy(), but your code will break when
ported to a system with different byte ordering.

If you're trying to guarantee network byte ordering, there are
functions that will handle this for you ("ntohl" and friends), but
they're not part of standard C. If your system has them, searching
your documentation for "ntohl" should be enough to get you started.

 

[Barry Schwarz]

Hi,

I have an array as follows -

char arr[100];

Now I wish to copy the following int -

int tmp = 0x01020304;

into this array, at element[12].

This following would have the same result -

ThePacket3[12] = 1;
ThePacket3[13] = 2;
ThePacket3[14] = 3;
ThePacket3[15] = 4;

How can I do this without breaking up my int into 4 bytes and assigning
each element one by one?

You could probably try this:

memcpy(&arr[12],&tmp,sizeof(int));

Just make sure you know exactly what you're doing. For example int
doesn't need to be 4 byte long.

And make sure that an int on the system is big-endian otherwise the
result will not be as specified.


Remove del for email

 

[Barry Schwarz]

int tmp = 0x01020304;

into this array, at element[12].

This following would have the same result -

ThePacket3[12] = 1;
ThePacket3[13] = 2;
ThePacket3[14] = 3;
ThePacket3[15] = 4;

How can I do this without breaking up my int into 4 bytes and assigning
each element one by one?

if you know that you're on a big-endian machine and sizeof char * 4 ==
sizeof int, you could theoretically do:

int *slice;
slice = &arr[12];
*slice = tmp;

You need a cast on the pointer assignment and the code will invoke
undefined behavior if arr[12] is not properly aligned for an int.

but that's obviously unportable as hell

Remove del for email

 

[CBFalconer]

I have an array as follows -

char arr[100];

Now I wish to copy the following int -

int tmp = 0x01020304;

into this array, at element[12].

This following would have the same result -

ThePacket3[12] = 1;
ThePacket3[13] = 2;
ThePacket3[14] = 3;
ThePacket3[15] = 4;

How can I do this without breaking up my int into 4 bytes and assigning
each element one by one?

Assuming your specs are accurate, and you change the type of tmp to
unsigned long, you can do it independant of endianism etc. by using
values, not bit patterns:

unsigned char ThePacket[100];
unsigned long tmp;
int i;

for (i = 3; i >= 0; i--, tmp /= 256)
ThePacket[12 + i] = tmp % 256;

You should be using unsigned items for both tmp and ThePacket. tmp
must be a long to guarantee 32 bits available.

If it can the compiler will probably optimize the modulo and
division operations into shifts and masks. Writing portable code
isn't all that hard, is it?

Notice that the values 3 and 12 above can be defined as FIELDSZ and
FIELDLOCN, or whatever nomenclature suits you.

 

[=?ISO-8859-1?Q?=22Nils_O=2E_Sel=E5sdal=22?=]

Hi,

I have an array as follows -

char arr[100];

Now I wish to copy the following int -

int tmp = 0x01020304;

into this array, at element[12].

This following would have the same result -

ThePacket3[12] = 1;
ThePacket3[13] = 2;
ThePacket3[14] = 3;
ThePacket3[15] = 4;

How can I do this without breaking up my int into 4 bytes and assigning
each element one by one?

Why that requirement ?
The obvious way is to bitshift to the right, mask
and assign the 4(assuming that's the size of your ints) bytes in the int
to the individual char array elements.