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swap variable values with pointer

so i have 2 variables x,y; pointer p1 stores x address and using p1 we make x value 99 then using p1 we point to y and make its value -300 then we create int temp and pointer p2 THIS IS THE PROBLEM: i need to swap values of variables x, y while using only existing pointers and int temp ( after swapping it should be x=-300; y=99) this is what i have so far: https://code.sololearn.com/cs6l8leG224j/#cpp

21st Dec 2019, 9:29 AM
A S
A S - avatar
5 ответов
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thing is, at this point (before swapping) p1 holds y's address/value, somehow i need to access x's address/value without using x variable
21st Dec 2019, 9:38 AM
A S
A S - avatar
0
it says in the original post that p2 is not assigned to any variable
21st Dec 2019, 9:44 AM
A S
A S - avatar
0
you can do it using a single pointer and a temp variable or just using two pointers.
21st Dec 2019, 12:18 PM
rodwynnejones
rodwynnejones - avatar
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using one pointer and a temp variable:- int main(){ int x, y, temp, *p1; p1 = &x; *p1 = 99; p1 = &y; *p1 = -300; cout <<"Before swap " << x << " " << y << endl ; temp = *p1; *p1 = x; x = temp; y = *p1; cout << "After swap " << x << " " << y; return 0; }
21st Dec 2019, 4:50 PM
rodwynnejones
rodwynnejones - avatar
0
using pointers only:- int main() { int x,y; int *p1=&x; *p1 = 99; int *p2 =&y; *p2 = -300; cout << "Before swap" << x << " " << y << endl; *p1 = *p1 + *p2 ; *p2 = *p1 - *p2; *p1 = *p1 - *p2; cout << "After swap" << x << " " << y; }
21st Dec 2019, 4:54 PM
rodwynnejones
rodwynnejones - avatar