What does "int* p=+s;" do?

Question

I saw a weird type of program here.

int main() {     int s[]={3,6,9,12,18};     int* p=+s; } 

Above program tested on GCC and Clang compilers and working fine on both compilers.

I curious to know, What does int* p=+s; do?

Is array s decayed to pointer type?

Answer 1

Built-in operator+ could take pointer type as its operand, so passing the array s to it causes array-to-pointer conversion and then the pointer int* is returned. That means you might use +s individually to get the pointer. (For this case it's superfluous; without operator+ it'll also decay to pointer and then assigned to p.)

(emphasis mine)

The built-in unary plus operator returns the value of its operand. The only situation where it is not a no-op is when the operand has integral type or unscoped enumeration type, which is changed by integral promotion, e.g, it converts char to int or if the operand is subject to lvalue-to-rvalue, array-to-pointer, or function-to-pointer conversion.

Answer 2

Test this:

#include <stdio.h> int main(){     char s[] = { 'h', 'e', 'l', 'l', 'o' , ' ', 'w', 'o', 'r', 'l', 'd', '!'} ;     printf("sizeof(s) : %zu,  sizeof(+s) : %zu\n", sizeof(s), sizeof(+s) ) ; } 

On my PC (Ubuntu x86-64) it prints:

sizeof(s): 12,  sizeof(+s) : 8 

where

12 = number of elements s times size of char, or size of whole array  8 = size of pointer