"taking address of temporary" warning

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taking address of temporary while accessing address of a element in vector<bool>

I get 'taking address of temporary' warning at line 7 in following piece of code

vector<bool> boolVect;
bool x = true;
boolVect.push_back(true);
boolVect.push_back(false);
ofstream fMetaOut("tmp.txt", ios::out);
fMetaOut.write((char* )&x, sizeof(bool));
fMetaOut.write((char* )&boolVect[0], sizeof(bool));
fMetaOut.close();

Why do I get this warning at line 7 and not at 6 ?? Both use boolean address.
How can I get rid of this warning?

Didn't you have this same issue [five months ago](http://stackoverflow.com/questions/8324699/taking-address-of-temporary-while-accessing-address-of-a-element-in-vectorbool)? — Blastfurnace, May 10 '12 at 05:16
@Blastfurnace there's no way you remembered that... is there? Tell me it's a coincidence you came across his old question! — Mahmoud Al-Qudsi, May 10 '12 at 05:27
@MahmoudAl-Qudsi: Well, simply click on Rahul's user name, you'll see all his previous questions :) — Gui13, May 10 '12 at 05:32
@MahmoudAl-Qudsi: I have a habit of looking at people's profiles and the third question down had the same warning message in the title. — Blastfurnace, May 10 '12 at 05:32

Alok Save · Answer 1 · 2012-05-10T05:26:25.787

x is not an temporary, it is a l-value, an automatic variable which resides on stack.
While, &boolVect[0] is not an l-value.
l-value is an entity whose address can be taken, in short a variable whose lifetime is long enough to reside in the memory so that it can be addressed by an name.

What is problem with vector<bool>?? It works fine for vector<int>

The C++ standard library provides a specialization of vector for boolean types. The goal is optimization, to use less size than a usual implementation of vector for type bool. A usual implementation for bool would reserve at least 1 byte for each element. The vector<bool> specialization usually uses internally only 1 bit for an element, so it is typically eight times smaller. But this optimization comes with a price:
In C++, the smallest addressable value must have a size of at least 1 byte. Thus, such a specialization of a vector needs special handling for references and iterators. As a result, a vector<bool> does not meet all requirements of other vectors. The behavior you see is one such example, elements of the vector<bool> are not a true l-values.

What is problem with vector?? It works fine for vector. – Rahul May 10 '12 at 05:15 — Rahul, May 10 '12 at 05:15
@Rahul: Ellaborated to explain in detail. – Alok Save May 10 '12 at 05:24 — Alok Save, May 10 '12 at 05:24

score 2 · Answer 2 · answered May 10 '12 at 05:23

Unlike other specialisations of vector, vector<bool> does not manage an array of bool objects. Instead, it manages a packed bitfield.

Since the bits are not individually addressable, boolVect[0] is not a reference to a bool; it's a (temporary) object of a type that's convertible to bool. There is no way to get a reference or pointer to a vector element.

If you want a container of boolean flags that you can take references to, then vector<char> or deque<bool> might be suitable choices.

Taking the address of x is fine, since it is a (non-temporary) bool object.

"taking address of temporary" warning

2 Answers2