调用非void函数而不使用其返回值。实际发生了什么？ [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

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，当你打电话给你时，会发生什么？非空函数返回一个对象，但你从来没有分配或使用所返回的对象？因此，不管你是否可以，因为我绝对知道你可以和理解上面的其他问题...编译器/运行时环境做什么？

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?

这不是一个特定于语言的问题，但如果您回答，请指定您要引用的语言，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

p>我相信对于C＃和Java，结果最终在堆栈上，然后编译器强制pop指令忽略它。 Eric Lippert的博文空洞是不变的有关于此的一些更多信息。

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，考虑以下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();
    }
}

由MS C＃4编译器生成的IL Main 方法是：

The IL generated (by the MS C# 4 compiler) for the Main method is:

.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 
}

注意对 pop 的调用 - 如果你使 Foo 为void方法就会消失。

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

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