Evaluation order of operands

Consider the below program.

C++

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// C++ implementation
#include <bits/stdc++.h>
using namespace std;
int x = 0;
  
int f1()
{
    x = 5;
    return x;
}
  
int f2()
{
    x = 10;
    return x;
}
  
int main()
{
    int p = f1() + f2();
    cout << ("%d ", x);
    getchar();
    return 0;
}

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C

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#include <stdio.h>
int x = 0;
  
int f1()
{
    x = 5;
    return x;
}
  
int f2()
{
    x = 10;
    return x;
}
  
int main()
{
    int p = f1() + f2();
    printf("%d ", x);
    getchar();
    return 0;
}

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Java

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class GFG {
  
    static int x = 0;
  
    static int f1()
    {
        x = 5;
        return x;
    }
  
    static int f2()
    {
        x = 10;
        return x;
    }
  
    public static void main(String[] args)
    {
        int p = f1() + f2();
        System.out.printf("%d ", x);
    }
}
  
// This code is contributed by Rajput-Ji

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Python3

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# Python3 implementation of the above approach 
class A():
    x = 0;
  
    def f1():
        A.x = 5;
        return A.x;
  
    def f2():
        A.x = 10;
        return A.x;
          
# Driver Code
p = A.f1() + A.f2();
print(A.x);
  
# This code is contributed by mits

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C#

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// C# implementation of the above approach
using System;
  
class GFG {
  
    static int x = 0;
  
    static int f1()
    {
        x = 5;
        return x;
    }
  
    static int f2()
    {
        x = 10;
        return x;
    }
  
    // Driver code
    public static void Main(String[] args)
    {
        int p = f1() + f2();
        Console.WriteLine("{0} ", x);
    }
}
  
// This code has been contributed
// by 29AjayKumar

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PHP

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<?php
// PHP implementation of the above approach 
$x = 0;
  
function f1()
{
    global $x;
    $x = 5;
    return $x;
}
  
function f2()
{
    global $x;
    $x = 10;
    return $x;
}
  
// Driver Code
$p = f1() + f2();
print($x);
  
// This code is contributed by mits
?>

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Output:

10

What would the output of the above program – ‘5’ or ’10’?
The output is undefined as the order of evaluation of f1() + f2() is not mandated by standard. The compiler is free to first call either f1() or f2(). Only when equal level precedence operators appear in an expression, the associativity comes into picture. For example, f1() + f2() + f3() will be considered as (f1() + f2()) + f3(). But among first pair, which function (the operand) evaluated first is not defined by the standard.



Thanks to Venki for suggesting the solution.

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