Why is sizeof() implemented as an operator in C++?

sizeof is not a real operator in C++. It is merely special syntax that inserts a continuing equal to the size of the argument. sizeof doesn’t want or have any runtime support. Sizeof cannot be overloaded because built-in operations, such as incrementing a pointer into an array implicitly depends on it.

The C standard specifies that sizeof should be implemented as an operator. In most compilers, the value of sizeof is substituted by a constant equal to it at the compile time itself.

example

#include <iostream>
using namespace std;
int main() {
   cout << "Size of char : " << sizeof(char) << endl;
   cout << "Size of int : " << sizeof(int) << endl;
   cout << "Size of short int : " << sizeof(short int) << endl;
   cout << "Size of long int : " << sizeof(long int) << endl;
   cout << "Size of float : " << sizeof(float) << endl;
   cout << "Size of double : " << sizeof(double) << endl;
   cout << "Size of wchar_t : " << sizeof(wchar_t) << endl;
   return 0;
}

Output

This will give the output −

Size of char : 1
Size of int : 4
Size of short int : 2
Size of long int : 4
Size of float : 4
Size of double : 8
Size of wchar_t : 4

Nancy Den

Published on 15-Feb-2018 12:53:19

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