Problem already solved, missed another warning in front. Thanks anyway. On 24-10-2006 10:33:48 +0200, Stefan Manegold wrote:
Just try with both
void myfunc(unsigned char *s) {
and
void myfunc(signed char *s) {
and you find out, whether you compiler be default defines char as signed or unsigned (and whether your systems/compilers header files define NULL as signed or unsigend) ;-)
Stefan
On Tue, Oct 24, 2006 at 10:16:11AM +0200, Fabian Groffen wrote:
I just figured that the following yields in a signed/unsigned warning:
1 void myfunc(char *s) { 2 if (s != NULL) { 3 ... 4 } 5 }
The error occurs on line 2. I cannot find the definition of NULL, but I suspect it is defined like: #define NULL 0
Obviously 0 is signed while s (pointer) is not. Maybe we should use #define NULL (char*)0 instead?
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