Why sizeof() method is giving different results?

Question

When I run the program below in C, I get the output result to be 4.

#include <stdio.h>

//using namespace std;

int main()
{
    printf("%d", sizeof('a')); 
  return 0; 
}

But when I run the code below in C++, I get the output result to be 1.

#include <iostream>

using namespace std;

int main()
{
    printf("%d", sizeof('a')); 
  return 0; 
}

Could you please explain why do I get different output for the same code as if 'a' is the way we define characters in both the languages ?

*why do I get different output for the same code* -- C and C++ are two different languages, even though they are "almost" identical. There are many other things that work one way in C++, but are different in C. — PaulMcKenzie, Mar 03 '20 at 16:15
Yeah, I know that. But I want to know the reason behind that. — Kundan, Mar 03 '20 at 16:16
@Kundan See: https://stackoverflow.com/questions/433895/why-are-c-character-literals-ints-instead-of-chars — NathanOliver, Mar 03 '20 at 16:23

Sourav Ghosh · Accepted Answer · 2020-03-03T16:10:51.857

0

In C, a character representation (like 'a') has type int. So, sizeof operator returns the size of an integer.

In C++, it's of a character type.

edited Mar 03 '20 at 16:10

answered Mar 03 '20 at 16:10

Sourav Ghosh

8

please vote to close obvious dupes. – NathanOliver Mar 03 '20 at 16:10
@Kundan: character literals such as `'a' ` have a data type just like numeric literals such as `1` and `2.3` have data types. The type of character literals in C is `int` while in C++, it's `char` – machine_1 Mar 03 '20 at 16:26
1

@machine_1 C calls `'a'` a _constant_ - not a _literal_. C's literals (like `"a"`) can have their address taken, Cannot do `&'a'` in C. – chux - Reinstate Monica Mar 03 '20 at 16:42

1 Answers1