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 <p>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.</p>
<p>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.</p>
<p>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.</p>
<p>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:</p>
<pre class="brush:php;toolbar:false"><code class="cpp">std::string str = getString();
const char* cStr = str.c_str();</code>

or:

<code class="cpp">const std::string& 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 </code>

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