Question

For this code:

#include<stdio.h>

void hello() { printf("hello\n"); }
void bye()   { printf("bye\n");   }

int main() {
    printf("%p\n", hello); 
    printf("%p\n", bye);
    return 0;
}

output on my machine:

0x80483f4
0x8048408

[second address is bigger in value]

on Codepad

0x8048541
0x8048511

[second address is smaller in value]

Does this have anything to do with endian-ness of the machines? If not,

• Why the difference in the ordering of the addresses?

• Also, Why the difference in the difference?

  0x8048541 – 0x8048511 = 0x30

  0x8048408 – 0x80483f4 = 0x14

---

Btw, I just checked. This code (taken from here) says that both the machines are Little-Endian

#include<stdio.h>
int main() {    
    int num = 1; 
    if(*(char *)&num == 1)
        printf("Little-Endian\n");
    else    
        printf("Big-Endian\n");
    return 0;       
}

Answer 1

It has nothing to do with endinanness, but with the C++ standard. C++ isn’t required to write functions in the order you see them to disk (and think about cross-file linking and even linking other libraries, that’s just not feasable), it can write them in any order it wishes.

About the difference between the actual values, one compiler might add guards around a block to prevent memory overrides (or other related stuff, usually only in debug mode). And there’s nothing preventing the compiler from writing other functions between your 2 functions. Keep in mind even a simple hello world application comes with thousands of bytes of executable code.

The bottom line is: never assume anything about how things are positioned in memory. Your assumptions will almost always be wrong. And why even assume? There’s nothing to be gained over writing normal, safe, structured code anyway.