Why my pointer becomes dangling pointer in c++?

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My main language is C#, and I'm learning opengl with scarce c++ background.

// readShaderSource returns const string

// no warning
auto vShaderStr = readShaderSource("vshader.glsl");
auto vShaderSource = vShaderStr.c_str();

// dangling pointer warning
auto vShaderSource = readShaderSource("vshader.glsl").c_str();

What I thought about dangling pointer is something like this:

int* a()
{
    auto a = 10;
    return &a

    // a dies here, so &a becomes dangling pointer
}

which does not seem to be the case.

Why should I have to go through string variable? Isn't it okay to directly access member function? Is there something to do with return value optimization? I'm pretty confused...

c++
1 Answers

What I thought about dangling pointer is something like this ... which does not seem to be the case.

It is the same case, actually. Just in a different form.

The pointer returned by std::string::c_str() is valid only for as long as the std::string object remains alive (and unmodified).

readShaderSource() returns a temporary std::string object. That temporary goes out of scope and is destroyed at the end of the full statement that calls readShaderSource().

So, in this code:

auto vShaderSource = readShaderSource("vshader.glsl").c_str();

The temporary std::string object goes out of scope and is destroyed on the ; after c_str() exits. Thus you are left with a dangling pointer to freed memory.

Whereas in this code:

auto vShaderStr = readShaderSource("vshader.glsl");
auto vShaderSource = vShaderStr.c_str();

You are saving the temporary std::string to a local variable, and then the temporary is destroyed on the ; after readShaderSource() exits. You are then calling c_str() on that local variable, so the returned pointer will remain valid for as long as that local variable remains in scope.

user_658630
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