Here are a few title options that fit the question-answer format and accurately describe the article\'s content: **Option 1 (Short & Direct):** * **Why Does Calling `std::string.c_str()` on a Re

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

Consider the following code:

<code class="cpp">std::string getString() {
    std::string str("hello");
    return str;
}

int main() {
    const char* cStr = getString().c_str();
    std::cout << cStr << std::endl; // Outputs garbage
}</code>

Intuitively, one would expect the returned temporary string from getString() to remain valid in main()'s scope. However, this is incorrect.

The issue lies in the lifetime of temporary objects in C . A temporary object is destroyed at the end of the expression in which it's created, unless it's bound to a reference or used to initialize a named object. In this case, getString() returns a temporary string, which is destroyed at the end of the expression in main().

Consequently, cStr holds a dangling pointer, and using it can lead to undefined behavior. To avoid this problem, one can use a named variable or reference to ensure the validity of the returned string. For instance:

<code class="cpp">std::string returnedString = getString();
const char* cStr = returnedString.c_str();
std::cout << cStr << std::endl; // Safe</code>

Alternatively, the temporary string can be used directly without assigning it to a variable:

<code class="cpp">std::cout << getString().c_str() << std::endl; // Also safe</code>

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