Please someone tell me why the output of var is 100 and not 20?? 😮

Question

#include <iostream>
using namespace std;

void myFunc(int *x) {
*x = 100;
}

int main() {
int var = 20;
myFunc(&var);
cout << var;
}

Answer 1

reference -> passes the address of variable, any changes to the variable in myfunc affects the variable in main
eg :
void fun(int *c)
{ c= 200;}

main ()
{ int x= 3;
fun(&x);
cout << x ; //output 200
}

value -> passes a copy of value as parameter
eg:
void fun(int c)
{ c= 200;}

main ()
{ int x= 3;
fun(x);
cout << x ; //output 3
}

Answer 2

originaly the value of var is 20, then you call myFunc passing as reference the memory direction of var, so now x points to var. Inside of the function you assign 100 to the value of x (x points to var) so var = 100 , then you print var which value is 100

Answer 3

x is passed to myFunc by reference, not by value.

Answer 4

https://www.sololearn.com/learn/CPlusPlus/1643/

Answer 5

Sreejith thank you alot it really helped

Answer 6

Damyan Petkov thank you but whats the main difference between passing by reference and by Value?