c++ why const& can get const values?

Question

I understand why int& x=1 is illegal(you can't have reference to constant value), but I don't understand why const int& x=1 is legal? How can you have reference to the number "1"? It is not even a defined variable.

EDIT: I read the answer given in this post: What happens when I assign a temporary int to a const reference in C++? but can someone explain what does he means "The lifetime of the temporary value returned by the expression f(1) will have its lifetime extended. This rule is unique for const references."? so literals are stored as int on the stack or not? Thanks!

I think it's an effect of const references extending temporaries' lifetime. — Quentin, Jul 21 '14 at 11:25

score 3 · Accepted Answer · answered Jul 21 '14 at 11:24

3

(you can't have reference to constant value)

You almost got it.

You cannot have a non-constant reference to a constant value.

You can have a constant reference to a constant value. (Which extends the lifetime of said value to the lifetime of that reference).

answered Jul 21 '14 at 11:24

Bartek Banachewicz

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The missing step in your answer is how a literal `1` turns into a value you can refer to. Which was the actual question. – Quentin Jul 21 '14 at 11:29
@Quentin Ah well, I think it's a duplicate then alright. – Bartek Banachewicz Jul 21 '14 at 11:30

score 1 · Answer 2 · answered Jul 21 '14 at 11:25

1

The second case with const reference is allowed, because then a temporary object of type int is created, and the reference refers to this temporary object.

answered Jul 21 '14 at 11:25

Wojtek Surowka

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Nasser Al-Shawwa · Answer 3 · 2014-07-21T11:54:05.977

This is a C++ language feature. Click here for an easy explanation.

EDIT

From the link:

Normally, a temporary object lasts only until the end of the full expression in which it appears. However, C++ deliberately specifies that binding a temporary object to a reference to const on the stack lengthens the lifetime of the temporary to the lifetime of the reference itself, and thus avoids what would otherwise be a common dangling-reference error

Ashalynd · Answer 4 · 2014-07-21T11:31:47.443

0

In this case you have the constant reference to a constant value, which under the hood means creating a temp variable whole life time is bound to the life time of the reference.

edited Jul 21 '14 at 11:31

answered Jul 21 '14 at 11:26

Ashalynd

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By default literals are not stored anywhere, so the second part of your answer is wrong. – Quentin Jul 21 '14 at 11:27
A literal in this case is a temp variable, which lifetime is bound to the lifetime of the reference. I will clarify this in the answer. – Ashalynd Jul 21 '14 at 11:30
And that's the actual answer to OP ;) – Quentin Jul 21 '14 at 11:31

score -1 · Answer 5 · answered Jul 21 '14 at 11:26

-1

Because C++ says so. The language is defined this way. The language creators decided to allow it.

answered Jul 21 '14 at 11:26

Lightness Races in Orbit

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c++ why const& can get const values?

5 Answers5