Confusing behavior of sizeof with chars [duplicate]

Question

#include <stdio.h> #include <string.h>  int main(void) {     char ch='a';      printf("sizeof(ch)          = %d\n", sizeof(ch));     printf("sizeof('a')         = %d\n", sizeof('a'));     printf("sizeof('a'+'b'+'C') = %d\n", sizeof('a'+'b'+'C'));     printf("sizeof(\"a\")       = %d\n", sizeof("a")); } 

This program uses sizeof to calculate sizes. Why is the size of 'a' different from the size of ch (where ch='a')?

sizeof(ch)          = 1 sizeof('a')         = 4 sizeof('a'+'b'+'C') = 4 sizeof("a")         = 2 

Answer 1

TL;DR - sizeof works on the type of the operand.

• sizeof(ch) == sizeof (char)-------------------(1)
• sizeof('a') == sizeof(int) --------------------(2)
• sizeof ('a'+ 'b' + 'c') == sizeof(int) ---(3)
• sizeof ("a") == sizeof (char [2]) ----------(4)

Let's see each case now.

1. ch is defined to be of char type, so , pretty straightforward.

2. In C, sizeof('a') is the same as sizeof (int), as a character constant has type integer.

   Quoting C11,

   An integer character constant has type int. [...]

   In C++, a character literal has type char.

3. sizeof is a compile-time operator (except when the operand is a VLA), so the type of the expression is used. As earlier , all the integer character constants are of type int, so int + int + int produces int. So the type of the operand is taken as int.

4. "a" is an array of two chars, 'a' and 0 (null-terminator) (no, it does not decay to pointer to the first element of the array type), hence the size is the same as of an array with two char elements.

---

That said, finally, sizeof produces a result of type size_t, so you must use %zu format specifier to print the result.