Why sizeof() operator gives different value for 'a' and "a" in C?

Question

As far as I know a character constant, viz, 'a', is stored in ASCII format which is internally treated as integer, 97 in case of 'a', that's why sizeof('a') returns 4 on executing but when I use sizeof("a") it returns 2. I have not found any explanation regarding that yet.

My code:

#include <stdio.h>
void main()
{
   int x,y;

   x = sizeof('a');
   y = sizeof("a");

   printf("%d\n",x);

   printf("%d",y);
}

which gives the output:

4
2

16-bit pointers? Wow, what are you learning to code on, a teapot? — Quentin, Oct 05 '18 at 13:50
@Quentin It's not a pointer, it's `const char[2]` (in C++, tags were changed). — Baum mit Augen, Oct 05 '18 at 13:51
**C** and **C++** are distinct languages and **this** is one of the questions where it actually matters. — Antti Haapala -- Слава Україні, Oct 05 '18 at 13:51
In C, `sizeof('a') == sizeof(int)` will always be true. `sizeof('a') == 4` will not always be true, and `'a' == 97` won't always be true either (there _are_ still quite a lot of IBM mainframes running on EBCDIC). — zwol, Oct 05 '18 at 13:53
The second one is a literal array of characters, including the finishing zero. Therefore, sizeof("a") is 2 (it is equivalent to { 'a', 0 }). The first one looks like a char type definition, but actually defaults to int size. Try: char a = 'a'; printf("%d\n", sizeof(a)); — KJH, Oct 05 '18 at 13:58
@BaummitAugen: the type of `"a"` in C is `/*readonly*/char[2]` (not `const char[2]`) -- don't know about C++. — pmg, Oct 05 '18 at 13:58
@pmg Wasn't sure about the C situation, that's why I included the annotation. It's `const char[2]` in C++. — Baum mit Augen, Oct 05 '18 at 14:01

Blaze · Accepted Answer · 2018-10-05T14:04:17.473

4

'a' is an integer. It has a size of 4 on most computers. It may also be something else, however. 2 is also common on more specialized hardware.

"a" is a string literal. There are two characters: a and \0. They have a size of 1 each, for a total of size 2. However, when you try to assign that, what you get usually is a const char* and that may have a different size.

edited Oct 05 '18 at 14:04

answered Oct 05 '18 at 13:50

Blaze

16,736
2
25
44

Wow, what a mixup. Thankfully, @PaulOgilvie already edited it for me. – Blaze Oct 05 '18 at 13:53
3

Were the downvotes because of the `\n` instead of `\0`? This answer is correct. – Kevin Oct 05 '18 at 13:55
1

@JamesSnook you don't add the size of all elements of an array to get the size of the array? That's what was meant by adding the size of the 2 characters together. – Kevin Oct 05 '18 at 13:59
1

@JamesSnook, could you tell us how then to explain this size 2? – Paul Ogilvie Oct 05 '18 at 13:59
1

@Kein my bad, it's been too long since I've used C, just wrote a test to double check and I was wrong. – James Snook Oct 05 '18 at 14:02
1

`"a"` has type `/*readonly*/char[2]`. In most contexts it is converted to a value of type `/*readonly*/char*` (not `const char*`). – pmg Oct 05 '18 at 14:03
1

@pmg Odd that I forgot this as I used to do `int len = sizeof(array) / sizeof(arrayType)` quite often. – James Snook Oct 05 '18 at 14:11

Why sizeof() operator gives different value for 'a' and "a" in C?

1 Answers1