memory - Please explain this result please. printf("%c", 'abcd')

Question

#include <stdio.h>

int main()
{
    printf("%c
", 'abcd');
    printf("%p
", 'abcd');
    printf("%c
", 0x61626364);
    printf("%c
", 0x61626363);
    printf("%c
", 0x61626365);
    return 0;
}

I want to ask this line : printf("%c ", 'abcd');
In this line, the result is 'd' but, I can't understand why 'd' is come out.
I tried to look other memories. In this situation, I found other memories have all alphabets.
Please explain me why result is 'd' and why other memories have all alphabets.
Thank you.

Answer 1

'abcd' is a multi-character constant, its value is implementation-defined.

C11 §6.4.4.4 Character constants section 10

An integer character constant has type int. The value of an integer character constant containing a single character that maps to a single-byte execution character is the numerical value of the representation of the mapped character interpreted as an integer. The value of an integer character constant containing more than one character (e.g., 'ab'), or containing a character or escape sequence that does not map to a single-byte execution character, is implementation-defined. If an integer character constant contains a single character or escape sequence, its value is the one that results when an object with type char whose value is that of the single character or escape sequence is converted to type int.

A common implementation gives 'abcd' a value of 'a' * 256 * 256 * 256 + 'b' * 256 * 256 + 'c' * 256 + 'd' (1633837924), you can check its value in your implementation by printing it using "%d". Although legal C, it's rarely used in practice.