In message <mailman.2231.1277700501.32709.python-l...@python.org>, Kushal
Kumaran wrote:
> On Sun, Jun 27, 2010 at 5:16 PM, Lawrence D'Oliveiro
> <l...@geek-central.gen.new_zealand> wrote:
>
>>In message <mailman.2184.1277626565.32709.python-l...@python.org>, Kushal
>> Kumaran wrote:
>>
>>> On Sun, Jun 27, 2010 at 9:47 AM, Lawrence D'Oliveiro
>>> <l...@geek-central.gen.new_zealand> wrote:
>>>
>>>> A long while ago I came up with this macro:
>>>>
>>>> #define Descr(v) &v, sizeof v
>>>>
>>>> making the correct version of the above become
>>>>
>>>> snprintf(Descr(buf), foo);
>>>
>>> Not quite right. If buf is a char array, as suggested by the use of
>>> sizeof, then you're not passing a char* to snprintf.
>>
>> What am I passing, then?
>
> Here's what gcc tells me (I declared buf as char buf[512]):
> sprintf.c:8: warning: passing argument 1 of ‘snprintf’ from
> incompatible pointer type
> /usr/include/stdio.h:363: note: expected ‘char * __restrict__’ but
> argument is of type ‘char (*)[512]’
>
> You just need to lose the & from the macro.
Why does this work, then:
l...@theon:hack> cat test.c
#include <stdio.h>
int main(int argc, char ** argv)
{
char buf[512];
const int a = 2, b = 3;
snprintf(&buf, sizeof buf, "%d + %d = %d\n", a, b, a + b);
fprintf(stdout, buf);
return
0;
} /*main*/
l...@theon:hack> ./test
2 + 3 = 5
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