How to show bytes of float

Question

I personally use a function show_bytes as follows:

#include<stdio.h>
typedef char *byte_pointer;

void show_bytes (byte_pointer x)
{
        int length = sizeof(float);
        int i;
        for(i = 0;i <length;i++)
        {
                printf("%2x",*(x+i));
                printf("\n");
        }
}

int main()
{
        float obj;
        printf("please input the value of obj:");
        scanf("%f",&obj);
        show_bytes((byte_pointer) &obj);
}

when i input 120.45,which should be 0x42f0e666

please input the value of obj:120.45
66
ffffffe6
fffffff0
42

why so many 'f' before the e6 and f0 while i use %.2x.

Answer 1

Your function should be:

void show_bytes (byte_pointer x)
{
   int i;
   for(i = 0; i <sizeof(float); i++)
   {
      printf("0x%2X\n", (unsigned int)(*(x++) & 0xFF));
   }
}

or

typedef uint8_t *byte_pointer;

void show_bytes (byte_pointer x)
{
   int i;
   for(i = 0; i <sizeof(float); i++)
   {
      printf("0x%2X\n", *(x++));
   }
}

In your code the problem is that the pointer type is signed an is promoted to signed int by printf.

%2X format does not limit the output digit, tells only to printf that the result string must be at least 2 characters long.

• First solution: the value is promoted to signed int but the passed value is truncated to the LSB.
• Second example: value is truncated by the type of pointer, that is unsigned char.

The rule is: to raw access memory, always use unsigned types.