Why does this version of the swap function fail to work? Is there a fix?

Question

```
void swap(int & lhs, int& rhs)
{
lhs = rhs;
rhs = lhs;
}
```

a) Of course it works! Just look at it. It clearly swaps the two parameters!
b) It fails because the programmer forgot to make the parameters call-by-reference.
c) It fails OK, and we can fix it we can just reverse the order of the lines.
d) It fails because the first line destroys the old value of lhs without saving it. Then both variables have the old value of rhs in them.
e) To fix this, we must save the lhs value in a local variable before making the first assignment indicated, then instead of the second line, assign the rhs the value of the local variable:
int local = lhs;
lhs = rhs;
rhs = local;

Answer 1

d) It fails because the first line destroys the old value of lhs without saving it. Then both variables have the old value of rhs in them., and e) To fix this, we must save the lhs value in a local variable before making the first assignment indicated, then instead of the second line, assign the rhs the value of the local variable:
int local = lhs;
lhs = rhs;

a) is argumentum ad baculum – argument by the stick. Shout it loud and it is the truth.. b) is incorrect, both parameters are call-by-reference. c) reversing the order of the assignments does not fix the fundamental problem of irretrievably destroying data.