什么都不捕获的 lambda 可以访问全局变量吗？

Question

int n;    
int main()
{
    [](){ n = 0; }(); // clang says "ok"

    int m;
    [](){ m = 0; }(); // clang says "not ok"
}

我只是想知道：

如果 lambda 不捕获任何内容，是否允许按照 C++ 标准访问全局变量？

Answer 1

是的，当然。适用正常名称查找规则。

[expr.prim.lambda]/7 ... for purposes of name lookup ... the compound-statement is considered in the context of the lambda-expression.

回复：为什么局部变量与全局变量的处理方式不同。

[expr.prim.lambda]/13 ... If a lambda-expression or an instantiation of the function call operator template of a generic lambda odr-uses (3.2) this or a variable with automatic storage duration from its reaching scope, that entity shall be captured by the lambda-expression.

[expr.prim.lambda]/9 A lambda-expression whose smallest enclosing scope is a block scope (3.3.3) is a local lambda expression... The reaching scope of a local lambda expression is the set of enclosing scopes up to and including the innermost enclosing function and its parameters.

在您的示例中，m 是一个从 lambda 的到达范围自动存储持续时间的变量，因此应被捕获。 n 不是，所以不一定是。

Answer 2

实际上 [](){ n = 10; }(); 不捕获任何东西，它使用全局变量代替。

int n;    
int main()
{
    [](){ n = 10; }(); // clang says "ok"
    std::cout << n; // output 10
}

见capture-list in Explaination

capture-list - a comma-separated list of zero or more captures, optionally beginning with a capture-default.

Capture list can be passed as follows (see below for the detailed description):

[a,&b] where a is captured by copy and b is captured by reference.

[this] captures the current object (*this) by reference

[&] captures all automatic variables used in the body of the lambda by reference and current object by reference if exists

[=] captures all automatic variables used in the body of the lambda by copy and current object by reference if exists

[ ] captures nothing

Answer 3

默认访问全局、静态和常量变量：

#include <iostream>

int n;    
int main()
{
    [](){ n = 10; }();
    std::cout << n << std::endl;
    static int m = 1;
    [](){ m = 100; }();
    std::cout << m << std::endl;
    const int l = 200;
    [](){ std::cout << l << std::endl; }();
}

什么都不捕获的 lambda 可以访问全局变量吗？

Can a lambda capturing nothing access global variables?

c++

lambda

standards

global-variables

c++11