assigning more than one character in char

Question

Why this program gives output \'y\'

#include 

int main(void) {
    char ch=\'abcdefghijklmnopqrstuvwxy\';
    printf(\"%c\",ch);         


        

Answer 1

If you intended to print out abcdefghijklmnopqrstuvwxy, then you should have stored it into a string variable instead of a char one (char ch[50] = char abcdefghijklmnopqrstuvwxy;).

String variables can hold more than one character, where as a char variable is for holding one character.