CODEWITHSUNDEEP
c++pointers
vfsoraki
vfsoraki

Reputation: 2307

regarding pointers in c++

i have a function like this

void some_func(some_type *var)
{
    // do something a get a var2 of type some_type*
    var = var2;
}

and in main i have

some_type *var;
some_func(var);
print_var(var);

but i get a segmentation fault when i run the program. debugging shows that print_var is the problem, as if var has no value. var2 is initialized using new some_type, so i think i don't need to do this for var.

but when i do var = new some_type in main, and i manually clone (copy each data of var2 to var) in some_func, i don't get the error.

where is my problem? i am not really used to pointer and memory allocation things, but i think both ways should work. am i wrong?

maybe the main question is when i allocate a pointer using new in a function, and then hold its address in an argument of pointer, does the memory get deallocated when the function finishes?

Upvotes: 0

Views: 95

Answers (3)

Uchia Itachi
Uchia Itachi

Reputation: 5315

Your someType *var in main() is local to main and the *var in the function is local to that function.

So, the var in function is pointing to var2 but not the one which is present in main.

Upvotes: 1

Tyson Jacobs
Tyson Jacobs

Reputation: 362

You are passing var by value. This means that a copy of var is made on the stack, separate from the var in main. This copy is then initialized with var2. But, when some_func exits, the stack is popped, and the copy is lost. The var declared in main is unchanged.

To fix this, one possible way would be for some_func to return a pointer value, which you can then assign to var.

Alternately, change your declaration of some_func to

void some_func(some_type *& var)

This passes var by reference, meaning that, for all you are concerned, the var in some_func is the same as the var in main.

Upvotes: 1

edwinc
edwinc

Reputation: 1676

You need to change your definition to:

    void some_func(some_type*& var)
    {
        // var will also be changed on the caller side
        var = var2; 
    }

Notice that it passes a reference to a pointer so the value of the pointer can be updated by the callee.

Upvotes: 3

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