Question

If I have the following code:

{
    UnicodeString sFish = L"FISH";
    char *szFish = AnsiString(sFish).c_str();

    CallFunc(szFish);
}

Then what is the scope of the temporary AnsiString that’s created, and for how long is szFish pointing to valid data? Will it still be valid for the CallFunc function?

Will it’s scope last just the one line, or for the whole block?

Answer 1

The C++11 standard $12.2.3 says:

When an implementation introduces a temporary object of a class that
has a non-trivial constructor (12.1, 12.8), it shall ensure that a
constructor is called for the temporary object. Similarly, the
destructor shall be called for a temporary with a non-trivial
destructor (12.4). Temporary objects are destroyed as the last step in
evaluating the full-expression (1.9) that (lexically) contains the
point where they were created. This is true even if that evaluation
ends in throwing an exception. The value computations and side effects
of destroying a temporary object are associated only with the
full-expression, not with any specific subexpression.

^{(emphasis mine)}

There are additional caveats to this, but they don’t apply in this situation. In your case the full expression is the indicated part of this statement:

char *szFish = AnsiString(sFish).c_str();
//             ^^^^^^^^^^^^^^^^^^^^^^^^^

So, the instant szFish is assigned, the destructor of your temporary object (i.e. AnsiString(sFish)) will be called and its internal memory representation (where c_str() points to) will be released. Thus, szFish will be immediately become a dangling pointer and any access will fail.

You can get around this by saying

CallFunc(AnsiString(sFish).c_str());

instead, as here, the temporary will be destroyed (again) after the full expression (that is, right at the ;) and CallFunc will be able to read the raw string.