Shortcut to remember the Order of Evaluation and Precedence of Operators in C [duplicate]

Answer 1

In C, the precedence table specify the order of evaluation of expression and also specify the association rules. Using that rule we can evaluate the equal priority operator(R->L OR L->R) in a expression.
You specify,
a = 1;
x = a++ + ++a;
1: ++a then exp: a(2)++ + 2 = 4 assign to x
2: and then increment a , becomes a = 3

suppose, a = a++ + ++a; then
increment a
2 + 2 assign to a(4). increment a(5).

Answer 2

Promod answer or his explanations is not correct

The correct method to remember is

Of all the pre operators -pre increment has highest priority

And of all the post increment operations -post increment has least priority

Addition has lower priority than both pre and post increment operators

Now consider the fact that Pre increment falls in the category of Right to left

And Post increment falls in the class of Left to right

Taking the two cases into consideration one can work out the code

i = 1;
i=i++ + ++i + ++i + i++;

To be not 14 but 12

Middle two expressions are evaluated to 4 then 4+4+4

"Postfix increment/decrement have high precedence, but the actual increment or decrement of the operand is delayed (to be accomplished sometime before the statement completes execution). "

This is the proof link enter link description here

Answer 3

If you find it confusing then so will anyone reading your code. If in doubt, use brackets to emphasize.

Answer 4

it is simple calculation When you write int a=1; x= a++ + ++a; Due to prefix increment operator (in ++a) the value of 'a' will become equal to 2 Therefore the current expression is equivalent to x= 2+2; // x=4 As soon as control goes to next line value of a increases by 1 due to postfix operator, now a =3

Now examine this statement int a=1; a= a++ + ++a; In this statement as explained above value of 'a' will be equal to the expression a= 2+2;. // means a=4 But due to postfix increment operator (a++) the value of a increases by 1 as soon as control shift to next line in the program. Therefore printf (%d,a); prints 5 I hope this will clear your doubt