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Why Does `c_str()` Work on a Deallocated `std::string` in This Case?

Published on 2024-11-11
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 Why Does `c_str()` Work on a Deallocated `std::string` in This Case?

Understanding Implicit Return in a C Function

In C , when passing a string literal to a function, the compiler implicitly converts it to a std::string object. However, in the following example, we return a C-style string literal from a std::string function:

std::string myFunction() {
    return "it's me!!";
}

This raises a concern because the std::string constructor implicitly called here creates a copy of the string literal. When the function returns, this copy should be deallocated, leaving a dangling pointer.

What Happens When You Call c_str()?

However, calling myFunction().c_str() returns a pointer to the data stored in the std::string object. This pointer points to the same memory that held the string literal, even after the std::string object is deallocated.

Why This Works (Sort of)

The reason this code appears to work is due to a quirk of the operating system's memory management. When a memory block is deallocated, the OS doesn't always clear its contents. This means that the string literal's data is still present in memory, even though it's technically unreachable.

Undefined Behavior and Luck

It's important to note that this behavior is undefined according to the C standard. This means that anything can happen, including crashes or incorrect results. It works in some cases not because of proper C practices, but because of the OS's implementation details.

Therefore, it's crucial to avoid relying on this behavior and always ensure that data is correctly allocated and deallocated in your C code.

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