Get value from memory address

Question

I have an int* variable that stores memory address, for the example address 0x28c1150.

I want to know, what value is stored at the address.

edit:

struct list {
    int value;
    list *next;
    list *head = NULL;
    void push(int n);
    void select();
    void pop();
    void top();

};

void list::push(int value) {
    list *temp = new list;

    temp->value = value;
    temp->next = head;
    head = temp;
}
void list::top(){
    list * temp = new list;

    cout << head;
}

i want to print top of my list

Or for the value that is being pointed to: `std::cout << *variable << "\n";` — Blaze, Mar 20 '19 at 07:38
You can't have a value without also having a *type*. What type of value are you expecting? — john, Mar 20 '19 at 07:38
@NoobXDDD Then `cout << *variable << '\n';` Does that not work for you? — john, Mar 20 '19 at 07:44
Possible duplicate of [cpp / c++ get pointer value or depointerize pointer](https://stackoverflow.com/questions/14420257/cpp-c-get-pointer-value-or-depointerize-pointer) — t.niese, Mar 20 '19 at 07:44
@NoobXDDD You realise you've completely changed the question? Why didn't you supply this information *at the start*. Do you think we have psychic powers? — john, Mar 20 '19 at 07:47
Please read ["What is the XY Problem?"](https://meta.stackexchange.com/questions/66377/what-is-the-xy-problem). Every little bit of information and edit you add shows more and more that's what you asked here. — StoryTeller - Unslander Monica, Mar 20 '19 at 07:47
And how is `int*` related to your code. I don't see any `int*` in your code. — t.niese, Mar 20 '19 at 07:48
Your `top()` function leaks memory. You allocate a new list and then just forget about it. — Ted Lyngmo, Mar 20 '19 at 07:51
no match for operator << (operand type are std::ostream {aka std::basic_ostream}' and list — NoobXDDD, Mar 20 '19 at 07:53
@NoobXDDD You are trying answers to the question everyone **thought** you asked. Try this instead `cout << head->value;` — john, Mar 20 '19 at 07:54

Ted Lyngmo · Answer 1 · 2019-03-20T07:53:42.777

1

If your variable is a list*:

list* variable = new list;
variable->top();

...but note that your current top() function leaks memory since you allocate a new list every time it's called and you just forget about it. Try this instead:

int list::top(){  
    return head->value;
}

std::cout << variable->top() << "\n";

edited Mar 20 '19 at 07:53

answered Mar 20 '19 at 07:45

Ted Lyngmo

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Thanks a lot that works fine, but can you tell me how to use this value in my code (in if to compare something) – NoobXDDD Mar 20 '19 at 09:52
`int other_value = 10; if(variable->top() == other_value) ...` ? I think you should edit your original question to make it clear what you are actually asking about. Right now it's put on hold. – Ted Lyngmo Mar 20 '19 at 11:17

score 0 · Answer 2 · answered Mar 20 '19 at 07:45

You have to dereference the pointer:

template<class T>
void print_value_at(T* pointer) {
    T& value = *pointer; //get value in pointer
    // print value 
    std::cout << value <<std::endl;
}

If the pointer is void, you have to cast it to whatever type it was originally:

int x = 10;
// get void pointer to x
void* x_pointer_as_void = (void*)&x; 
// convert it back to a pointer to an int:
int* x_pointer = (int*)x_pointer_as_void;

score 0 · Answer 3 · answered Mar 20 '19 at 07:50

0

Here it is (based on current version of question)

cout << head->value;

answered Mar 20 '19 at 07:50

john

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Get value from memory address

3 Answers3