Lambda capture: to use the initializer or not to use it?

问题

Consider the following, minimal example:

int main() {
    int x = 10;    
    auto f1 = [x](){ };
    auto f2 = [x = x](){};
}

I've seen more than once such an use of the initializer [x = x], but I can't fully understand it and why I should use it instead of [x].
I can get the meaning of something like [&x = x] or [x = x + 1] (as shown in the documentation and why they differ from [x], of course, but still I can't figure out the differences between the lambdas in the example.

Are they fully interchangeable or is there any difference I can't see?

回答1:

There are various corner cases that pretty much boils down to "[x = x] decays; [x] doesn't".

• capturing a reference to function:

  void (&f)() = /* ...*/;
  [f]{};     // the lambda stores a reference to function.
  [f = f]{}; // the lambda stores a function pointer
  

• capturing an array:

  int a[2]={};
  [a]{}     // the lambda stores an array of two ints, copied from 'a'
  [a = a]{} // the lambda stores an int*
  

• capturing a cv-qualified thing:

  const int i = 0; 
  [i]() mutable { i = 1; } // error; the data member is of type const int
  [i = i]() mutable { i = 1; } // OK; the data member's type is int
  

来源：https://stackoverflow.com/questions/36188694/lambda-capture-to-use-the-initializer-or-not-to-use-it