#include <stdio.h> int main() { printf(5 + "Good Morning\n"); return 0; } The code prints Morning. Should the code print Morning or should it show undefined behavior?
2 Answers
It should show 'Morning'.
You are using pointer arithmetic - though you appear not to know it! "Good Morning\n" is a char * pointer to a constant string. You are then adding 5 to this pointer, which advances it by 5 characters. Hence the pointer now points to the 'M' of 'Morning'.
Comments
The code is correct since printf is defined as:
int printf ( const char * format, ... ); And according to pointers arithmitic 5 + "Good Morning\n" is a pointer to the first element of "Morning\n". So the statment:
printf(5 + "Good Morning\n"); has the same result as:
printf("Morning\n"); Explanation:
|G|o|o|d| |M|o|r|n|i|n|g|\n| ^ ^ | | "Good Morning\n" >---- | + | 5 >---------------------- 
1 Comment
abligh
It's not entirely equivalent as the binary will still contain the whole string. This might be significant if the function, unlike
printf, backtracked to bytes prior to the pointer passed - as well as the fact it wastes space in normal circumstances.
printf("%s", "Good Morning\n" + 5)instead