## Why Does `std::string.c_str()` Fail When Used on a Function Returning a String in C ?

Understanding Why std::string.c_str() on a Function Returning a String Fails

In C , when a function returns a string, it does so by value. This means that the returned string is a copy of the original string, rather than a reference to it. This is different from the behavior of other built-in types, such as integers or floats, which are returned by reference by default.

In your example, the function getString() returns a string object that is a copy of the string "hello." This copy is stored in a temporary location in memory. The line const char* cStr = getString().c_str(); attempts to get a C-style string representing the returned string. However, because the temporary string has already been destroyed, the pointer cStr now points to invalid memory.

The Destruction of Temporary Objects

The explanation for this behavior lies in the rules for temporary objects in C . A temporary object is an object that is created implicitly by the compiler to avoid unnecessary copying. In your case, the temporary string created by getString() is a temporary object.

Temporary objects are destroyed at the end of the full expression in which they are created. A full expression is an expression that is not a subexpression of some other expression. In your example, the line const char* cStr = getString().c_str(); is a full expression. Therefore, the temporary string created by getString() is destroyed at the end of this line.

Preventing the Issue

To prevent this issue, you have several options:

• Store the returned string in a named variable before extracting a C-style string.
• Bind the returned string to a const lvalue-reference or rvalue-reference.
• Use the C-style string pointer before the temporary string gets destroyed.

By following these guidelines, you can avoid the problems associated with dangling pointers and ensure that your code behaves correctly.

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