Post-increment and Pre-increment concept?

int i, x;

i = 2;
x = ++i;
// now i = 3, x = 3

i = 2;
x = i++; 
// now i = 3, x = 2

'Post' means after - that is, the increment is done after the variable is read. 'Pre' means before - so the variable value is incremented first, then used in the expression.

The difference between the postfix increment, x++, and the prefix increment, ++x, is precisely in how the two operators evaluate their operands. The postfix increment conceptually copies the operand in memory, increments the original operand and finally yields the value of the copy. I think this is best illustrated by implementing the operator in code:

int operator ++ (int& n)  // postfix increment
{
    int tmp = n;
    n = n + 1;
    return tmp;
}

The above code will not compile because you can't re-define operators for primitive types. The compiler also can't tell here we're defining a postfix operator rather than prefix, but let's pretend this is correct and valid C++. You can see that the postfix operator indeed acts on its operand, but it returns the old value prior to the increment, so the result of the expression x++ is the value prior to the increment. x, however, is incremented.

The prefix increment increments its operand as well, but it yields the value of the operand after the increment:

int& operator ++ (int& n)
{
    n = n + 1;
    return n;
}

This means that the expression ++x evaluates to the value of x after the increment.

It's easy to think that the expression ++x is therefore equivalent to the assignmnet (x=x+1). This is not precisely so, however, because an increment is an operation that can mean different things in different contexts. In the case of a simple primitive integer, indeed ++x is substitutable for (x=x+1). But in the case of a class-type, such as an iterator of a linked list, a prefix increment of the iterator most definitely does not mean "adding one to the object".

All four answers so far are incorrect, in that they assert a specific order of events.

Believing that "urban legend" has led many a novice (and professional) astray, to wit, the endless stream of questions about Undefined Behavior in expressions.

So.

For the built-in C++ prefix operator,

++x

increments x and produces (as the expression's result) x as an lvalue, while

x++

increments x and produces (as the expression's result) the original value of x.

In particular, for x++ there is no no time ordering implied for the increment and production of original value of x. The compiler is free to emit machine code that produces the original value of x, e.g. it might be present in some register, and that delays the increment until the end of the expression (next sequence point).

Folks who incorrectly believe the increment must come first, and they are many, often conclude from that certain expressions must have well defined effect, when they actually have Undefined Behavior.