CODEWITHSUNDEEP
c++referencelocal
linuxfever
linuxfever

Reputation: 3823

C++: Local reference

I would like to understand what happens in the following code

struct A
{
  vector<double> x;
};

void f(A &a)
{
  vector<double> &y = a.x;
}

When the function f exits, is a.x destroyed? Thanks in advance!

Upvotes: 1

Views: 1070

Answers (3)

shawn
shawn

Reputation: 336

In this case, when you enter f() you locally create a reference y to a.x that exists independently from a and the rest of the world. When you leave f() the locally created reference y goes out of scope and gets destroyed. The rest of the world stays as it was before you entered f().

Upvotes: 3

Kerrek SB
Kerrek SB

Reputation: 476930

Remember that references are not objects. y is a variable, but not an object. It is a reference to the existing object a.x, but that object itself is not local to the scope of f. So the variable y goes out of scope at the end of f, but the object to which it refers does not.

Upvotes: 4

mathematician1975
mathematician1975

Reputation: 21351

No a.x is not destroyed. You simply have created a local reference to a.x and then the function exits - nothing is changed. Your code effectively does nothing at all.

Upvotes: 4

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