sizeof on string literal defined in #define VS char*

Consider the following:

Code:

---

#define STR "some long string"
 
int main(int argc, char* argv[]) {
    char c* = "some long string";
    printf("%d\n", sizeof(c))
    printf("%d\n", sizeof(STR));
    return 0;
}

---

This prints out:
8
17
What’s the explanation for the difference?
Shouldn't I expect the same result in both cases?

Use strlen instead of sizeof to get length of C-strings (zero byte terminated char arrays).

Note: The size returned does not count the zero byte at the end.

Tim S.

Recall that c is a pointer, whereas STR is an array of char including the terminating null character.

Also compare with

Code:

---

    char c[] = "some long string";
    char c[100] = "some long string";

---

It should also be mentioned that it is incorrect to use "%d" to print the return value of the sizeof operator unless you also explicitly cast the return value to an int. sizeof returns size_t which is not only unsigned but can be 64 bits even when int is 32 bits. If that's the case on your system then the only reason you get the correct output is because your machine is little endian so that the low 32-bits comes first.

So to print the value of sizeof you should either use the correct format spec, "%zu", or cast it to int:

Code:

---

#include <stdio.h>
#include <string.h>
 
int main() {
    char str[100] = "a string";  // Salem's example
 
    printf("zu: %zu\n", sizeof str);
    printf(" d: %d\n", (int)sizeof str);
 
    printf("\nstrlen: %zu\n", strlen(str));
   
    printf("\nNumber of bytes (chars) in:\n");
    printf("  int:    %zu\n", sizeof(int));
    printf("  size_t: %zu\n", sizeof(size_t));
 
    printf("\nCheck endianness:\n");
    int n = 1;
    printf("  %s endian\n", *(char*)&n ? "little" : "big");
 
    return 0;
}

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