为什么未捕获的变量会发出警告? [英] Why non-captured variable is raising warning?

问题描述

在VS 2015 中考虑以下代码:

Consider this code in VS2015:

int a,b;

[]
{
    int a;  //  C4456: declaration of 'a' hides previous local declaration
}; 

为什么lambda中的a发出这样的警告?在VS2013中可以正常编译.

Why a in lambda giving such warning? It compiles fine in VS2013.

编辑:有趣的是(并且错误地)，以下内容在VS2013中不是错误:

Interestingly, (and incorrectly), following is not an error in VS2013:

 [a]
 {        
     int a; // No error, even if `a` is captured.
     a++;
 };

推荐答案

第一个警告肯定看起来像是编译器错误.

The first warning definitely looks like a compiler bug.

第二个不是bug，因为您在不同的范围内声明了它.该变量仅被捕获，捕获未声明.

The second one isn't a bug since you're declaring it in a different scope. The variable is only captured, it is not declared by the capture.

考虑可能会生成的功能对象

Think about the function object this could generate

class foo {
  foo(int a): a(a) {}

  void operator()() {
    int a;
  }

  int a;
};

a的两个声明之间没有冲突，并且由于lambda编译为类似这样的内容，所以这就是捕获不关心内部声明的原因.

There's no conflict between the two declarations of a, and since the lambda compiles to something like this, that's why the capture doesn't care about the inner declaration.

更新:这与

void foo(int a) {
  int a;
}

因为使用lambda，它将被编译为带有operator()的类，并且捕获将作为构造函数参数传入，这就是为什么它们位于不同范围的原因.

because in the case of the lambda, it will get compiled to a class with an operator(), and the captures will get passed in as constructor parameters, which is why they're in a different scope.

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