What are applications of pre-increment and post-increment operators why these operators behaviour are made like this is there any reason for such behaviour?

Answer 1

All operators must return a value, and it is this return value (the evaluation of the operation) that differentiates the pre-increment and post-increment operators. If the return value is of no concern, then the pre-increment operator is the preferred version, particularly when working with objects. This is because the post-increment operator's implementation will generally make a copy of the original object which it will return after incrementing the original object. If the return value is of no concern, then making a copy adds an expensive overhead (this doesn't apply to primitive data types since CPU optimisation can pre-empt the return value without requiring a copy to be made).

Applications for post-increment and pre-increment are many and varied, but ultimately the goal is to increment the value. Whether you use post-increment or pre-increment is largely down to whether you need to use the return value or not, and whether you need to use the original value or the incremented value.

Although the same thing can be achieved with longhand code, it is more efficient to use the appropriate post-increment or pre-increment operators.

For instance:

int x=1;

int y=x;

++x;

Can be implemented more efficiently with:

int x=1;

int y= x++;

Whereas:

int x=1;

++x;

int y=x;

Can be implemented more efficiently with:

int x=1;

int y= ++x;

Use caution when pre-incrementing the same value in a compound statement. For example:

int x=0;

int y= ++x * ++x;

The expected result is 2, but the actual result is 4. This is because both expressions (++x) return the same value: a reference to x. And since x is incremented twice, the expression evaluates to 2*2, not 1*2.

To get the expected result of 2, each operation must be evaluated separately:

int x=0;

int y= ++x;

y*= ++x;