C++ pointer creation and assignment inside the function

Question

What is the difference between *p_address= new int(2) and the assignment via & p_address = &value considering that both examples take place inside the function? For example: I've got the int pointer *original_pointer. I pass its address to the function. Inside the function I create an int pointer which points to the int value of 2. Then I assign the pointer (which is created inside the function) to the *original_pointer. When I cout the *original_pointer outside the function, it returns -858993460, while inside the function it returns the value of 2. However, when I use new to create a pointer inside the function, the value of *original_pointer inside and outside of the function is the same. Here is the code:

int main() {
    while (true) {
        void assign_(const int**);
        char* tmp = " ";
        int const *original_pointer;
        assign_(&original_pointer);
        cout << "the address of original_pointer is " << original_pointer << endl;
        cout << "the value of original_pointer is " << *original_pointer << endl;
        cin >> tmp;
    }
    return 0;
}
void assign_( int const **addr) {
    int* p_value;
    int value = 2;
    p_value = &value;
    *addr = p_value;
    //*addr = new RtFloat(2.0);   // If I create the pointer this way the value of *addr is the same with *original_pointer
    cout << "the adress of *addr inside the function is " << *addr << endl;
    cout << "the value of **addr inside the function is " << **addr << endl;
}

John3136 · Answer 1 · 2015-05-20T00:39:53.557

1

*p_address= new int(2) Allocates memory for ~~2 ints~~ one int (with the value 2) which "lives" until you delete it.

p_address = &value just sets p_address to the address of a local variable which becomes invalid as soon as the function exits (as you've seen).

edited May 20 '15 at 00:39

answered May 19 '15 at 06:44

John3136

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new int(2) doesn't allocate 2 ints. new int[2] does. – Benjy Kessler May 19 '15 at 06:46
@BenjyKessler D'oh. You're right. Will edit to correct. – John3136 May 20 '15 at 00:37

score 0 · Answer 2 · answered May 19 '15 at 06:44

0

The difference between *p_address= new int(2) and the assignment via &p_address = &value; Is that in the first case the value pointed to by p_address is on the heap and in the second case it is on the stack. This means that the lifetime of the value pointed to by p_address in the first case is until it is deleted and in the second case it is only alive until value goes out of scope.

answered May 19 '15 at 06:44

Benjy Kessler

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*p_address= new int(2) is a actually a dynamic memory allocation integers on the heap memory. The program creates memory on heap. This basically means that addresses will remain valid till 1. Delete[] operator is used explicitly by program.2. The process execution stops. When this happens the OS will reallocate the memory to some other process. Most modern OS do this. p_address = &value is basically assigning the address of a variable to the pointer. This gets allocated on the stack. This means the variable will only be valid as long as it is within a scope. – Spanky May 19 '15 at 06:59

score 0 · Answer 3 · answered May 19 '15 at 06:46

When you use int value = 2; it is created on the stack. Once the function returns the variable is automatically deallocated.

When you use *p_address= new int; the memory is allocated on the heap. it is not automatically deallocated. so even after the function returns that memory is there.

C++ pointer creation and assignment inside the function

3 Answers3