Print out the value uint8_t *

Question

The followings are my snippet:

typedef struct{
   uint8_t dstEndpt;
   uint16_t srcEndpt;
   uint8_t * quality;
} DataIndT;

DataIndT * dataIndPtr;

printf("dstEndpt: 0x%02x", dataIndPtr->dstEndpt); <-- print out the value of dstEndpt

printf("dstEndpt: 0x%04x", dataIndPtr->srcEndpt); <-- print out the value of srcEndpt

However, how can I print out the value of quality ?

Answer 1

However, how can I print out the value of quality ?

You do

printf("%p", (void*) dataIndPtr->quality);

This will print address, since value of pointer is address to object to which pointer points.

---

To print the object where the pointer points, in this case, you can use format specifiers available for C99 (also need to include inttypes.h). Of course you also need to dereference the pointer:

printf("%" PRIu8 "\n", *(dataIndPtr->quality));

since quality is pointer to uint8_t or

printf("%" PRIu16 "\n", *(dataIndPtr->srcEndpt));

for uint16_t types.

Answer 2

quality is a pointer, or like an array, if you want to print the value that points to you need to specify it. with the index or dereferencing it:

printf("quality: %d", *(dataIndPtr->quality));

Using the zero index like if it was an array should also print the value:

printf("quality: %d", dataIndPtr->quality[0]);

Or if what you want is print the value of the pointer itself then Michal's answer is what you want.

Answer 3

To print the value of a pointer, use %p:

printf("dstEndpt: %p", (void*)dataIndPtr->quality); 

Answer 4

Like @Giorgi pointed you can use inttypes.h.

Here is a small example:

#include <inttypes.h>
#include<stdio.h>

int main(void){

    uint8_t a = 0;
    uint16_t b = 0;
    uint32_t c = 0;
    uint64_t d = 0;


    printf("%" PRId8 "\n", a);
    printf("%" PRIu16 "\n",b);
    printf("%" PRIu32 "\n", c);
    printf("%" PRIu64 "\n", d);

    return 0;
}