How the memcmp on structure with integer variable in c lang compares. Result is not as expected

Question

I have a structure with integer, I am comparing the struct by using memcmp, I don't want to use other memthods.

   #include <stdio.h>
   #include <string.h>

   typedef struct Foo 
   {
      int d;
   } Foo;

   int main(int argc, const char * argv[])
   {
      Foo one, two;
      int result;
   
      memset(&one,0,sizeof(Foo));
      memset(&two,0,sizeof(Foo));
    
      one.d = 1022;
      two.d = 1024;
    
      result = memcmp((void*)&one, (void*)&two, sizeof(Foo));
      printf("comp with assignment %d\n",result);
    
      if (result == 0) printf("Arrays are the same\n");
    
      return 0;
   }

memcmpa should return -1, but it returns 1. why ? memcmp woth one.d = 1022 and two.d = 1023 will return correct value. why is so?

It compares _bytes_, not `int`s - look at how an `int` looks at byte level and you'll see. — Ted Lyngmo, Aug 19 '22 at 11:06
Look up `little endian` and `big endian`... Ted has pointed the way... — Fe2O3, Aug 19 '22 at 11:08
Another note: _"should return -1, but it returns 1"_ - It will return a negative, positive or `0`. Not `-1` , `1` or `0`. — Ted Lyngmo, Aug 19 '22 at 11:51

0___________ · Accepted Answer · 2022-08-19T11:22:22.077

If you add two printfs in your code:

   typedef struct Foo 
   {
      int d;
   } Foo;

   int main(int argc, const char * argv[])
   {
      Foo one, two;
      int result;
   
      memset(&one,0,sizeof(Foo));
      memset(&two,0,sizeof(Foo));
    
      one.d = 1022;
      two.d = 1024;

      printf("%04x %04x\n", one.d, two.d);
    
      result = memcmp((void*)&one, (void*)&two, sizeof(one));
      printf("comp with assignment %d\n",result);
    
      if (result == 0) printf("Arrays are the same\n");
    
      return 0;
   }

Result:

03fe 0400
comp with assignment 1

You will see that the first byte of one is 0xfe and the first byte of two is 0x00 (they are in opposite order as most modern machines are little endioan) So 0xfe > 0x00 and memcmp returns 1

score 0 · Answer 2 · answered Aug 19 '22 at 11:31

It compares bytes, not ints:

memcmp- This function reads object representations, not the object values, and is typically meaningful for byte arrays only: structs may have padding bytes whose values are indeterminate,

Look at how an int looks at byte level and you'll see it more clearly. It can be stored with the most significant byte first or last - and the result of memcmp will depend on that.

You can create your own memcmp_debug for this purpose.

Example:

int memcmp_debug(const void *vpa, const void *vpb, size_t len) {
    const unsigned char *a = vpa, *b = vpb;

    puts("comparing these:");
    for(size_t i = 0; i < len; ++i) {
        printf("%2d %02X %02X\n", i, a[i], b[i]);
    }

    puts("\ncomparing:");
    for(unsigned i = 0; i < len; ++i) {
        int result = (int)a[i] - (int)b[i];
        printf("%2d %02X %02X  =>  %d\n", i, a[i], b[i], result);
        if(result) return result;
    }
    return 0;
}

Possible output:

comparing these:
 0 FE 00
 1 03 04
 2 00 00
 3 00 00

comparing:
 0 FE 00  =>  254

.. and here it returned on the first byte compared (the least significant byte on my machine) and returned a positive value just like it did for you.

How the memcmp on structure with integer variable in c lang compares. Result is not as expected

2 Answers2