在递归中使用+ =在Java和C ++中有不同的结果 [英] Different results in Java and C++ using += in recursion

问题描述

下面的非常简单的 Java 代码具有奇怪的输出，但 C 和 C ++ 中的相同逻辑代码具有正确的输出。我试着用jdk1.7和jdk1.3（相对jre），奇怪的输出总是存在。我是来自 C 和 C ++ 世界的 Java 新手，现在我疯了。请帮帮我。任何答案将非常感激：

The very simple Java code as follows has the weird output, but the same logic code in C and C++ has the right output. I try with the jdk1.7 and jdk1.3(relative jre), the weird output is always there. I am a newbie in Java from C and C++ world and now I am going crazy. Please help me. Any answer will be highly appreciated:

public class Test {

    public static int sum=0;

    public static int fun(int n) {

        if (n == 1)
            return 1;
        else
            sum += fun(n - 1);  // this statement leads to weird output
        // { // the following block has right output
        //     int tmp = fun(n - 1);
        //     sum += tmp;
        // }

        return sum;
    }

    public static void main(String[] arg) {
        System.out.print(fun(5));
    }
}

输出为1， / C ++代码如下：

The output is 1 which should be 8. Relative C/C++ code is as follows:

#include<stdio.h>
int sum=0;
int fun(int n) {

        if (n == 1)
            return 1;
        else
            sum += fun(n - 1);

        return sum;
    }

int main()
{
    printf("%d",fun(5));

    return 0;
}

添加测试Java代码：

Adding test java code:

class A {
    public int sum = 0;

    public int fun(int n) {
        if(n == 1) {
            return 1;
        } else {
            sum += fun(n - 1);
            return sum;
        }
    }
}

public class Test {
    public static void main(String arg[]){
        A a = new A();
        System.out.print(a.fun(5));
    }
}

推荐答案

I为了给出一个完整的答案，要经历这个乐趣（3）。对于那些不感兴趣的人，为什么这对C ++而言，但不适用于Java，请忽略我的回答。

I'm going to run through this for fun(3) for the sake of giving a complete answer. For those of you who are not interested why this works for C++ but not for Java, please ignore my answer.

这里是Java在做什么： strong>

Here is what Java is doing:

里面fun（3）

sum += sum + fn(n-1) // sum is 0

成为

sum = 0 + fun(2) // sum is 0

then fun（2）

Then inside fun(2)

sum = 0 + fun(1) // sum is 0

然后在fun（1）

return 1 // sum is 0

< 2）

Back inside fun(2)

sum = 0 + 1; // sum is 0

成为

sum = 1; // sum will soon become 1

sum = 0 + 1; // sum is 1

成为

sum = 1; // sum gets reset to 1

C ++正在做什么：

里面fun（3）

sum += fn(n-1) // sum is 0

成为

sum = sum + fn(2) // sum is 0

$ b b

then fun（2）

Then inside fun(2)

sum = sum + fn(1) // sum is 0

然后在里面fun（1）

Then inside fun(1)

return 1 // sum is 0

p>

Back inside fun(2)

sum = sum + 1 // sum is 0

成为

sum = 0 + 1 => sum = 1 // sum will soon become 1

b
$ b

Back inside fun(3)

sum = sum + 1 // sum is 1

成为

sum = 1 + 1 // sum will soon become 2

你应该做什么：
我不知道为什么C ++求值 sum 在调用函数之后，而不是之前。我不知道这是否在规格。但我知道，你不应该依赖于这在任何语言。正确的解决方案是：

What you should do: I do not know why C++ evaluates sum after making the function call rather than before. I do not know if this is in the specifications. But I do know that you should not be depending on this in any language. A correct solution would be:

int fun(int n) {
    if (n == 1)
        return 1;
    else
        return n + f(n - 1);
}

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