Why does sizeof("string") prints one more than the length of "string"?

Question

# include <stdio.h>
int main()
{
        int a = sizeof("string");
        printf("%d", a );
        return 0;
}

The above code prints 7 as the output while the length of string is only 6. Could someone please explain?

maybe the hidden null character added to the end counts as a byte? Normally when you use sizeof, you use it on some kind of variable, not a fixed value. — Mike -- No longer here, Oct 27 '15 at 06:58
@Mike `sizeof` works equally well on variables, values, and types. For all three, it only evaluates the ___type___ of the expression (at compile-time). — Emil Laine, Oct 27 '15 at 07:14

score 11 · Answer 1 · answered Oct 27 '15 at 06:57

This is what is happening:

The string literal "string" is an array of (const) char, including a null-terminator character, i.e. {'s', 't', 'r', 'i', 'n', 'g', '\0'}. In your example, said array has 7 elements.
The operator sizeof when applied to an array yields its size in bytes.
The size of an array is the sum of the size of each of its elements.
The size of one char is 1.

So, you get the number of chars explicit in the literal, plus a null-terminator.

This behaviour is the same in both C and C++

Magisch · Answer 2 · 2015-10-27T07:19:29.497

All strings in c, even the ones you write like "string" are terminated with a null byte '\0'.

The expression "string" in c or c++ code is considered a string literal, that means a char array that looks like this: {'s','t','r','i','n','g','\0'}. In c and c++ all strings must be terminated by a null byte in memory, otherwise functions cannot identify their end.

This is, by the way, also why you must add 1 character when calculating lenght of character arrays in c code to account for the null byte. So if you want to store "string" in a character array that you declare, you would have to do char array[7]; using 6 here would lead to undefined behaviour.

score 2 · Answer 3 · answered Oct 27 '15 at 06:56

2

The length of the string is 6 and the null character is also added by default when we specify the string in "". So it prints 7.

answered Oct 27 '15 at 06:56

score 2 · Answer 4 · answered Oct 27 '15 at 06:57

2

Strings are null-terminated and stored as an array containing the characters and terminated with a null character ('\0', called NUL in ASCII). So you have:

s | t | r | i | n | g | \0 | 
                         ^-- 7th char

answered Oct 27 '15 at 06:57

Sadique

22,572
7
65
91

score 1 · Answer 5 · answered Oct 27 '15 at 06:56

1

Because "string" is implicitly adding a '\0' character.

answered Oct 27 '15 at 06:56

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Why does sizeof("string") prints one more than the length of "string"?

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