Simple printf with sizeof does not work at all.

Question

I got the most simple code to display sizeof() of a datatype, say an int.

#include <stdio.h>
int main() { 
  printf('%i', sizeof(int));
}

No matter what I do, such as put sizeof(int) into an integer, or use 'zu' instead of 'i', it hands me this error:

error: invalid conversion from ‘int’ to ‘const char*’

Is there something wrong with my compiler? I do not get why I cannot print such a simple sizeof..

EDIT: It seems a printf('%s', 'foo'); STILL tells me I am converting int to const char*, how on earth??

Answer 1

Try print("%i", sizeof(int));

double quotes is missing

Also explore if you can use cstdio instead of stdio.h if you are on C++.

Answer 2

Single quotes ' are used to delimit character literals (type int in C) while double quotes " are used to delimit string literals (type "array of char" in C).

printf is declared as follows:

int printf (const char * format, ...)

Thus you are attempting to pass an int where the function expects a const char * or something that can be converted to const char *.

Note: In C++ character literals are type char, string literals are "array of const char".

Answer 3

It should be:

printf("%i\n", sizeof(int));

It should be the double quotes.