CODEWITHSUNDEEP
c++visual-studio-2015lambda
Ajay
Ajay

Reputation: 18411

Why non-captured variable is raising warning?

Consider this code in VS2015:

int a,b;

[]
{
    int a;  //  C4456: declaration of 'a' hides previous local declaration
};

Why a in lambda giving such warning? It compiles fine in VS2013.

EDIT: Interestingly, (and incorrectly), following is not an error in VS2013:

 [a]
 {        
     int a; // No error, even if `a` is captured.
     a++;
 };

Upvotes: 2

Views: 362

Answers (1)

Jakub Arnold
Jakub Arnold

Reputation: 87210

The first warning definitely looks like a compiler bug.

The second one isn't a bug since you're declaring it in a different scope. The variable is only captured, it is not declared by the capture.

Think about the function object this could generate

class foo {
  foo(int a): a(a) {}

  void operator()() {
    int a;
  }

  int a;
};

There's no conflict between the two declarations of a, and since the lambda compiles to something like this, that's why the capture doesn't care about the inner declaration.

Update: This is something completely different from

void foo(int a) {
  int a;
}

because in the case of the lambda, it will get compiled to a class with an operator(), and the captures will get passed in as constructor parameters, which is why they're in a different scope.

Upvotes: 3

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