Why does pre-increment work but post-increment does not on a reference variable?
#include <iostream>
void swap(int&, int&);
int main()
{
int x=10, y=20;
int &a=x, &b=y;
swap(++a, ++b); //swap (a++,b++) is not allowed.
printf("%d %d ", a, b);
return 0;
}
void swap(int& x, int& y)
{
x+=2;
y+=3;
}
Why is swap(++a, ++b) allowed but swap(a++, b++) says:
[Error] invalid initialization of non-const reference of type 'int&' from an rvalue of type 'int'
When you call
swap (a++,b++)
a++ and b++ give you a temporary object since post incrementing returns the previous values. Since swap() takes its parameters by reference they cannot bind to those temporary values. using ++a and ++b work as we first increment a and b and then pass it to swap so there is no temporary.
++a returns an l-value but a++ returns an r-value. You need to pass an l-value to the swap() function.
Relevant post on l-valu and r-value: postfix (prefix) increment, L-value and R-value (in C and C++)
The function std::swap must swap the value of two variable, aka a lvalue reference. The expression x++ is not a variable, but a value, or a rvalue. A rvalue cannot be bound to a lvalue reference.
The difference between a rvalue and a lvalue could be explained that way:
int p;
/* p is left value, 3 is right value, ok */
p = 3;
But the following is not valid:
/* not ok, 3 is right value */
3 = p;
What you are sending to std::swap are two value of numbers.
Passing l-value / variable:
int x = 5, y = 9;
swap(x, y); // allowed bcos x and y are l-values / variables
pre-increment (also pre-decrement):
int x = 5, y = 9;
swap(++x, ++y); // allowed bcos ++x and ++y are l-values
Passing r-value / value:
swap(5, 9); // NOT allowed bcos 5 and 9 are values (NOT variables)
post-increment (also post-decrement):
int x = 5, y = 9;
swap(x++, y++); // NOT allowed bcos x++ and y++ are r-values
