c – 可以使用lambda捕获全局变量吗？

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

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

我只是好奇：

如果lambda没有捕获任何东西,是否允许根据C标准访问全局变量？

解决方法

是的,当然.正常名称查找规则适用.

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

Re：为什么局部变量与全局变量不同.

[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.

在你的例子中,m是一个变量,它具有从lambda到达范围的自动存储持续时间,因此将被捕获. n不是,所以不一定是.

原文地址：https://www.jb51.cc/c/111648.html