c++ - Passing a point in to a function -

in program i've created 2 pointers (a,b) points memory address of x , y. in function i've created supposed swap memory address of , b(so b=a , a=b). when compile gives me error (invalid conversion 'int' 'int*') mean? i'm passing pointer function or read regular int?

#include <iostream> using std::cin; using std::cout; using std::endl; void pointer(int* x,int* y)// swaps memory address a,b {     int *c;     *c = *x;     *x = *y;     *y = *c; }  int main() {     int x,y;     int* = &x;     int* b = &y;     cout<< "adress of a: "<<a<<" adress of b: "<<b<<endl; // display both memory address      pointer(*a,*b);     cout<< "adress of a: "<<a<<" adress of b: "<<b<<endl; // displays swap of memory address        return 0; } 

error message:

c++.cpp: in function 'int main()':
c++.cpp:20:16: error: invalid conversion 'int' 'int*' [-fpermissive]
c++.cpp:6:6: error: initializing argument 1 of 'void pointer(int*, int*)' [-fpermissive]
c++.cpp:20:16: error: invalid conversion 'int' 'int*' [-fpermissive]
c++.cpp:6:6: error: initializing argument 2 of 'void pointer(int*, int*)' [-fpermissive]

in function call

pointer(*a,*b); 

expressions *a , *b have type int while corresponding parameters of function have type int *.

if want swap 2 pointers , not values (objects x , y) pointed pointers function should following way

void pointer( int **x, int **y )// swaps memory address a,b {     int *c = *x;     *x = *y;     *y = c; } 

and called like

pointer( &a, &b ); 

or define parameters having reference types. example

void pointer( int * &x, int * &y )// swaps memory address a,b {     int *c = x;     x = y;     y = c; } 

and call like

pointer( a, b );