Why Does Calling `std::string.c_str()` on a Function Returning a String by Value Lead to Problems?

Why Calling std::string.c_str() on a Function That Returns a String May Lead to Problems

When dealing with strings in C , understanding the lifetime of temporary objects is crucial. A common misunderstanding occurs when calling std::string.c_str() on the return value of a function that returns a string by value.

In the following code snippet:

<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;
}</code>

One might expect that the function getString() would return a copy of the string str stored within the function scope, allowing cStr to hold a valid reference to it.

However, this is not the case. In C , the return value of a function call is a temporary object. When a temporary object is bound to a reference or used to initialize a named object, its lifetime is extended. Otherwise, it is destroyed at the end of the full expression in which it was created.

In the code above, the return value of getString() is a temporary object of type std::string, which is returned by value. Since cStr does not bind to a reference or initialize a named object, the temporary std::string object is destroyed after the expression getString() is evaluated. This means cStr now points to deallocated memory, making it unsafe to use.

To avoid this issue, consider extending the lifetime of the temporary object by copying it into a named object or binding it to a reference, such as:

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

or:

<code class="cpp">const std::string&amp; ref = getString();
const char* cStr = ref.c_str();</code>

Alternatively, you can use the pointer returned by c_str() immediately without storing it in a variable:

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

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