CODEWITHSUNDEEP
c++string
DalekSupreme
DalekSupreme

Reputation: 1513

Is string dynamically allocated?

I have got a quick question. I have the following code:

class Class1
{
    Class1();
    ~Class1();
    void func1();
private:
    char* c;

}

void Class1::func1()
{
    string s = "something";
    this->c = s.c_str();
}

will c store "something" when func1() finishes?

Upvotes: 3

Views: 180

Answers (4)

Christian Ternus
Christian Ternus

Reputation: 8492

The variable s, will go out of scope once control exits that block, at which point its destructor will be called.

When is an object "out of scope"?

Upvotes: 1

user529758
user529758

Reputation:

No. It will invoke undefined behavior instead. (if you dereference the pointer, anyway.) Since s is a block-scope object with automatic storage duration, it is destroyed when the function returns, and that renders the pointer returned by .c_str() invalid.

Why not use an std::string member variable instead?

Upvotes: 8

digital_revenant
digital_revenant

Reputation: 3324

s is a local variable of type std::string in Class::func1. Once func1() finishes, the string s will go out of scope.

Any pointers you have with the address of s stored in them will become dangling pointers.

Upvotes: 3

paddy
paddy

Reputation: 63471

It will store a dangling pointer that you must not access. It may contain the string "something" or it may not. It doesn't matter, because accessing it is undefined behaviour, and should be avoided completely.

If you want to copy the string do this:

c = strdup( c.c_str() );

And don't forget to free(c) in ~Class1()

Beware that if you call func1 twice, you will leak memory. You probably want to initialise c to NULL in the constructor, and call free(c) before reassigning it in func1.

Surely a better approach is to store a std::string instead of a char*, which manages memory properly for you.

Upvotes: 1

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