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* .

Answer 1

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 *.

Answer 2

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

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

Answer 3

The format specifier for printing void pointers using printf in C is %p. What usually gets printed is a hexadecimal representation of the pointer (although the standard says simply that it is an implementation defined character sequence defining a pointer).