调用非void函数不使用它的返回值。到底发生了什么？ [英] Calling a non-void function without using its return value. What actually happens?

问题描述

所以，我发现了一个类似的问题这里，但答案是更多的风格，你是否是能够做到这一点。

So, I found a similar question here, but the answers are more about style and whether or not you are able to do it.

我的问题是，当你调用一个实际发生的事情非void函数返回一个对象，但你永远不分配或使用表示返回的对象？所以，少谈你是否可以，因为我绝对知道你可以和理解上面链接的另一个问题......是什么编译/运行环境呢？

My question is, what actually happens when you call a non-void function that returns an object, but you never assign or use said returned object? So, less about whether or not you can, because I absolutely know you can and understand the other question linked above... what does the compiler/runtime environment do?

这是不是一种语言的具体问题，但如果你回答，请注明你指的是什么语言，因为行为会有所不同。

This is not a language specific question, but if you answer, please specify what language you are referring to, since behaviors will differ.

推荐答案

我相信，无论对于C＃和Java，结果在堆栈中结束，然后编译器会强制弹出指令忽略它。埃里克利珀对博客文章虚空是不变的< 。/ A>有一些这方面的更多信息。

I believe that for both C# and Java, the result ends up on the stack, and the compiler then forces a pop instruction to ignore it. Eric Lippert's blog post on "The void is invariant" has some more information on this.

例如，请考虑下面的C＃代码：

For example, consider the following C# code:

using System;

public class Test 
{
    static int Foo() { return 10; }

    public static void Main()
    {
        Foo();
        Foo();
    }
}

的

生成的IL（由MS C＃4编译）为主方法是：

.method public hidebysig static void Main() cil managed
{
    .entrypoint
    .maxstack 8
    L_0000: call int32 Test::Foo()
    L_0005: pop 
    L_0006: call int32 Test::Foo()
    L_000b: pop 
    L_000c: ret 
}

请注意来电来弹出 - 如果你让富一个void方法，它消失

Note the calls to pop - which disappear if you make Foo a void method.

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