Why does calling std::string.c_str() on a function that returns a string not work?

by

getString() would return a copy of str (getString() returns by value);

It’s right.

thus, the copy of str would stay “alive” in main() until main() returns.

No, the returned copy is a temporary std::string, which will be destroyed at the end of the statement in which it was created, i.e. before std::cout << cStr << std::endl;. Then cStr becomes dangled, dereference on it leads to UB, anything is possible.

You can copy the returned temporary to a named variable, or bind it to a const lvalue-reference or rvalue-reference (the lifetime of the temporary will be extended until the reference goes out of scope). Such as:

std::string s1 = getString();    // s1 will be copy initialized from the temporary
const char* cStr1 = s1.c_str();
std::cout << cStr1 << std::endl; // safe

const std::string& s2 = getString(); // lifetime of temporary will be extended when bound to a const lvalue-reference
const char* cStr2 = s2.c_str();
std::cout << cStr2 << std::endl; // safe

std::string&& s3 = getString();  // similar with above
const char* cStr3 = s3.c_str();
std::cout << cStr3 << std::endl; // safe

Or use the pointer before the temporary gets destroyed. e.g.

std::cout << getString().c_str() << std::endl;  // temporary gets destroyed after the full expression

Here is an explanation from [The.C++.Programming.Language.Special.Edition] 10.4.10 Temporary Objects [class.temp]]:

Unless bound to a reference or used to initialize a named object, a
temporary object is destroyed at the end of the full expression in
which it was created. A full expression is an expression that is
not a subexpression of some other expression.

The standard string class has a member function c_str() that
returns a C-style, zero-terminated array of characters (§3.5.1, §20.4.1). Also, the operator + is defined to mean string concatenation.
These are very useful facilities for strings . However, in combination they can cause obscure problems.
For example:

void f(string& s1, string& s2, string& s3)
{

    const char* cs = (s1 + s2).c_str();
    cout << cs ;
    if (strlen(cs=(s2+s3).c_str())<8 && cs[0]==´a´) {
        // cs used here
    }

}

Probably, your first reaction is “but don’t do that,” and I agree.
However, such code does get written, so it is worth knowing how it is
interpreted.

A temporary object of class string is created to hold s1 + s2 .
Next, a pointer to a C-style string is extracted from that object. Then
– at the end of the expression – the temporary object is deleted. Now,
where was the C-style string allocated? Probably as part of the
temporary object holding s1 + s2 , and that storage is not guaranteed
to exist after that temporary is destroyed. Consequently, cs points
to deallocated storage. The output operation cout << cs might work
as expected, but that would be sheer luck. A compiler can detect and
warn against many variants of this problem.

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