7.5 Postincrement and Postdecrement

Using ‘++’ or ‘--’ after an lvalue does something peculiar: it gets the value directly out of the lvalue and then increments or decrements it. Thus, the value of i++ is the same as the value of i, but i++ also increments i “a little later.” This is called postincrement or postdecrement.

For example,

#include <stdio.h>   /* Declares printf. */

int
main (void)
{
  int i = 5;
  printf ("%d\n", i);
  printf ("%d\n", i++);
  printf ("%d\n", i);
  return 0;
}

prints lines containing 5, again 5, and 6. The expression i++ has the value 5, which is the value of i at the time, but it increments i from 5 to 6 just a little later.

How much later is “just a little later”? The compiler has some flexibility in deciding that. The rule is that the increment has to happen by the next sequence point; in simple cases, that means by the end of the statement. See Sequence Points.

Regardless of precisely where the compiled code increments the value of i, the crucial thing is that the value of i++ is the value that i has before incrementing it.

If a unary operator precedes a postincrement or postincrement expression, the increment nests inside:

-a++   is equivalent to   -(a++)

That’s the only order that makes sense; -a is not an lvalue, so it can’t be incremented.

The most common use of postincrement is with arrays. Here’s an example of using postincrement to access one element of an array and advance the index for the next access. Compare this with the example avg_of_double (see An Example with Arrays), which is almost the same but doesn’t use postincrement.

double
avg_of_double_alt (int length, double input_data[])
{
  double sum = 0;
  int i;

  /* Fetch each element and add it into sum.  */
  for (i = 0; i < length;)
    /* Use the index i, then increment it.  */
    sum += input_data[i++];

  return sum / length;
}