Can a lambda capturing nothing access global variables?

Question

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

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

I just wonder:

Answer 1

Actually the [](){ n = 10; }(); doesn't capture anything, it uses the global variable instead.

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

See 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