Dave Robillard wrote:
>> Because char* usually means, you know, a pointer, not a variable length
>> array :)
>>
> .... char buf[] is, you know, equivalent to char* buf. You do know C,
> yes? ;)
>
You *do* know C, yes? Well enough to judge others?
Hint: try this little proggy (gcc should compile it fine):
#include <stdio.h>
int main(int argc, char *argv[])
{
struct X
{
int a;
char buf[];
};
struct Y
{
int a;
char *buf;
};
printf("%d %d\n", sizeof(struct X), sizeof(struct Y));
}
/* Krzysztof */
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Received on Fri Nov 30 04:15:04 2007
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