操作数的评估顺序

考虑下面的程序。

C++

// C++ implementation
#include 
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;
}

C

#include 
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;
}

Java

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

Python3

# 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

C#

// 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

Javascript

x = 0;
 
function f1()
{
    x = 5;
    return x;
}
 
function f2()
{
    x = 10;
    return x;
}
 
 
var p = f1() + f2();
document.write(x);
     
// This code is contributed by Amit Katiyar

输出：

10

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