Printing a void* variable in C

Question

Hi all I want to do a debug with printf. But I don't know how to print the "out" variable.

Before the return, I want to print this value, but its type is void* .

int  hexstr2raw(char *in, void *out) {     char c;     uint32_t i = 0;     uint8_t *b = (uint8_t*) out;     while ((c = in[i]) != '\0') {         uint8_t v;         if (c >= '0' && c <= '9') {             v = c - '0';         } else if (c >= 'A' && c <= 'F') {             v = 10 + c - 'A';         } else if (c >= 'a' || c <= 'f') {             v = 10 + c - 'a';         } else {             return -1;         }         if (i%2 == 0) {             b[i/2] = (v << 4);             printf("c='%c' \t v='%u' \t b[i/2]='%u' \t i='%u'\n", c,v ,b[i/2], i);}         else {             b[i/2] |= v;             printf("c='%c' \t v='%u' \t b[i/2]='%u' \t i='%u'\n", c,v ,b[i/2], i);}         i++;     }     printf("%s\n", out);     return i; } 

How can I do? Thanks.

Answer 1

printf("%p\n", out); 

is the correct way to print a (void*) pointer.

Answer 2

This:

uint8_t *b = (uint8_t*) out; 

implies that out is in fact a pointer to uint8_t, so perhaps you want to print the data that's actually there. Also note that you don't need to cast from void * in C, so the cast is really pointless.

The code seems to be doing hex to binary conversion, storing the results at out. You can print the i generated bytes by doing:

int j; for(j = 0; j < i; ++j)   printf("%02x\n", ((uint8_t*) out)[j]); 

The pointer value itself is rarely interesting, but you can print it with printf("%p\n", out);. The %p formatting specifier is for void *.