MESSAGE
| DATE | 2004-04-23 |
| FROM | Billy
|
| SUBJECT | Re: [hangout] Debugging cdparanoia in C
|
On Thu, Apr 22, 2004 at 09:16:38PM -0400, Ruben Safir Secretary NYLXS wrote:
> sizeof also works with structs and arrays.
> (a string is an array of chars)
> 1 #include
> 2 #include
> 3
> 4 int main(int argc,char **argv ){
> 5 char buffer[] = "My Dog Has Fleas!";
> 6 char * buf2;
> 7 int size = 0;
> 8 size = sizeof(buffer);
> 9 printf("Size of Buffer is %i\n", size);
> 10 printf("Buffer is %s\n", buffer);
> 11 buf2 = (char*)malloc(255);
> 12 size = sizeof(buf2);
> 13 printf("Size of Buffer is %i\n", size);
> 14 strncpy(buf2, buffer, 5);
> 15 size = sizeof(buf2);
> 16 printf("Size of Buffer is %i\n", size);
> 17 printf("Buffer is %s\n", buf2);
> 18 exit(1);
> 19 }
>
> ./a.out
>
> Size of Buffer is 18
> Buffer is My Dog Has Fleas!
> Size of Buffer is 4
> Size of Buffer is 4
> Buffer is My Do
>
>
> The puzzle here is that sizeof seems to not work with the char pointer.
The SIZE of the char pointer object 'buf2', which is 4 bytes.
The SIZE of the char array object 'buffer' is 18.
There is a lot of confusion out there between arrays and pointers.
It pays to learn the difference.
> I'm not happy about that. I wonder how it behaves with strings passed as
> params. It probibly doesn't.
There are no strings in C. Only char arrays and char pointers.
You pass pointers, not arrays. Don't expect too much of sizeof.
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