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？
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.
