Sizeof string literal

Question

The following code

#include <iostream>    
using namespace std;

int main()
{
    const char* const foo = "f";
    const char bar[] = "b";
    cout << "sizeof(string literal) = " << sizeof( "f" ) << endl;
    cout << "sizeof(const char* const) = " << sizeof( foo ) << endl;
    cout << "sizeof(const char[]) = " << sizeof( bar ) << endl;
}

outputs

sizeof(string literal) = 2
sizeof(const char* const) = 4
sizeof(const char[]) = 2

on a 32bit OS, compiled with GCC.

1. Why does sizeof calculate the length of (the space needed for) the string literal ?
2. Does the string literal have a different type (from char* or char[]) when given to sizeof ?

Answer 1

sizeof returns the size in bytes of its operand. That should answer question number 1. ;) Also, a string literal is of type "array to n const char" when passed to sizeof.

Your test cases, one by one:

• "f" is a string literal consisting of two characters, the character f and the terminating NULL.
• foo is a pointer (edit: regardless of qualifiers), and pointers seem to be 4 bytes long on your system..
• For bar the case is the same as "f".

Hope that helps.

Answer 2

1. sizeof("f") must return 2, one for the 'f' and one for the terminating '\0'.
2. sizeof(foo) returns 4 on a 32-bit machine and 8 on a 64-bit machine because foo is a pointer
3. sizeof(bar) returns 2 because bar is an array of two characters, the 'b' and the terminating '\0'.

The string literal has the type 'array of size N of char' where N includes the terminal null.

Remember, arrays do not decay to pointers when passed to sizeof.