Increment/Decrement Operators

Incrementing (adding 1 to) and decrementing (subtracting 1 from) a variable are so common that they have their own operators in C. There are actually two version of each operator — a prefix version and a postfix version.

Operator	Symbol	Form	Operation
Prefix increment	++	++x	Increment x, then evaluate x
Prefix decrement	––	––x	Decrement x, then evaluate x
Postfix increment	++	x++	Evaluate x, then increment x
Postfix decrement	––	x––	Evaluate x, then decrement x

The prefix increment/decrement operators are very straightforward. The value of x is incremented or decremented, and then x is evaluated. For example:

1 2	int x = 5; int y = ++x; // x is now equal to 6, and 6 is assigned to y

The postfix increment/decrement operators are a little more tricky. The compiler makes a temporary copy of x, increments x, and then evaluates the temporary copy of x.

1 2	int x = 5; int y = x++; // x is now equal to 6, and 5 is assigned to y

In the second line of the above example, x is incremented from 5 to 6, but y is assigned the value of the copy of x, which still has the original value of 5.

Here is another example showing the difference between the prefix and postfix versions:

1 2 3 4 5 6

int x = 5, y = 5; cout << x << " " << y << endl; cout << ++x << " " << --y << endl; // prefix cout << x << " " << y << endl; cout << x++ << " " << y-- << endl; // postfix cout << x << " " << y << endl;

This produces the output:

5 5
6 4
6 4
6 4
7 3

On the third line, x and y are incremented/decremented before they are evaluated, so their new values are printed by cout. On the fifth line, a temporary copy of the original values (x=6, y=4) is sent to cout, and then the original x and y are incremented. That is why the changes from the postfix operators don’t show up until the next line.