Pointers

Question

I understand that pointers point to a memory address, but what do they really do? can they access out of scope variables? ex:
why would i:
int x = 5;
int *ptr = &x;
cout << *ptr << endl;

and not:
int x = 5;
cout << x << endl;

Answer 1

Hazer C++ Begginer
You could, yes. But imagine you are working on a large code, thousands of lines. It is good practice to make your code modular, that is, isolate certain functionalities in functions. By writing functions you can reutilize code in several parts of a program without having to write it again.

This does not successfully swaps the values held in the variables:

void swap(a, b) {
int tmp = a;
a = b;
b = tmp;
}

If you try this in the main function:

int x = 3;
int y = 5;
swap(x, y);

Then 3 is still in "x" and 5 is still in "y". Because only the values of the variables were passed to the swap function, not the addresses.

Answer 2

Hazer C++ Begginer you use pointers when you need to pass variables by reference and to manipulate the data in specific memory locations, and in general that's it.
I think the reason why people stress over pointers is because it is a concept that takes some time to fully "sink in".

Answer 3

You can pass variables by reference when using pointers.

A classic example is a function to swap the values of two variables.

void swap(int *a, int *b) {
int tmp = *a;
*a = *b;
*b = tmp;
}

This function successfully swaps the values in memory, because the addresses of the variables are passed to it.

int x = 3;
int y = 5;
swap(&x, &y);

Now 5 is stored in "x" and 3 is in "y".

Answer 4

so when i call a function like this it changes the return variable to whatever i type when i call it? I just dont know how it knew that you were talking about temp when you called it Ketan Lalcheta

Answer 5

Hazer C++ Begginer Not necessarily only for "organization". That's just the case for the swap function. Passing by reference means you are altering the values in the variables which are outside of the scope of the function.

I recommend you read more on passing by reference: http://www.learncpp.com/cpp-tutorial/73-passing-arguments-by-reference/

Answer 6

Hazer C++ Begginer pointer is a life saver when you deal with big amount of data...

void print(int x)
{
cout << x << endl;
}

void print(int* x)
{
cout << *x << endl;
}

consider above example.. you are using pointer in second case while normal variable in first case...

if you call function having pointer as argument, new copy of variable is not created and memory address of variable itself (which you declared in main function) is passed to function...

where as in normal function, new variable is created which is local to that function only... compiler has to create a copy of variable in function and need to destroy the same at the end of function call.

so, effectively you are creating less number of variable by directly accessing same variable through address of variable using pointer...

this can be done using reference as well in c++ but reference has some different limitations compared to pointer..

Answer 7

Ketan Lalcheta How does that work? you arent even modifying the temp variable?

Answer 8

Hazer C++ Begginer in this scenario, you will not appreciate importance of function returning reference... sometimes, it is must to return reference... case I recall for same is << operator overloading for object cout

Answer 9

I don't understand this pointer lesson .can enybody define that lesson to me please..

Answer 10

Hazer C++ Begginer making it global violates oops concepts... any one can access it from any class... you should wrap your data and functions to avoid data being accessed by unauthorised access..

that's the reason we use encapsulation concept... data as private or protected and member function to access those data as public...

Answer 11

Hazer C++ Begginer

getvariable() = "Sololearn"; is replaced as temp = "Sololearn"; as function returns variable (not value)

Answer 12

Hazer C++ Begginer I think I got your confusion... let me give you another better example:

Use below code and observe output that temp1 is updated after function call:

string temp1 = "Test1";
string temp2 = "Test2";

string& getVariable()
{
return temp1 ;
}

int main()
{

//before function call
cout << "temp1 is " << temp1 << endl;
cout << "temp2 is " << temp2 << endl;

getVariable() = "Updated";

//after function call
cout << "temp1 is " << temp1 << endl;
cout << "temp2 is " << temp2 << endl;

return 0;
}

Now change a line in getvalue function with return temp2 instead of temp1...
observe output now and you may see that temp2 value gets updated..

Answer 13

so it changes the return value? would that be the same as taking a parameter and changing the string to that parameter? Ketan Lalcheta

Answer 14

as in normal function, we display int x = getsum();
cout << x;
displays value returned by function, reference returns variable from function...

Answer 15

i think i fixed it Ketan Lalcheta

Answer 16

but if it does the same thing, why use one over the other? Ketan Lalcheta

Answer 17

Someone,can u explain to me what is the real purpose of the pointer????

Answer 18

I don't understand about the location memory and all

Answer 19

but couldnt i just do

int main() {
int x = 5;
int y = 10;
int tmp = x;
x = y;
y = tmp;

}

Answer 20

but isnt

string temp;
string &getstring() {
return temp;
}
int main() {
getstring() = "test";
return 0;
}

the same as

string getstring(string input) {
return input;
}

int main() {
getstring("hello");
return 0;
}