CONFUSED over OPERATOR PRECEDENCE

[dakuneko]

Assuming that a, b and c are of integer type whose initial value is 0:

	(1)  a = ++b + ++c;
	(2)  a = b++ + c++;
	(3)  a = ++b + c++;
	(4)  a = b-- + --c;

The result are as follows:

	(1) 2  
	(2) 2  
	(3) 5  
	(4) 5

Now I am confused. It seems that in (2) a = b++ + c++;, the + gets evaluated first before the ++.

How is this so? From what I've read, ++ has a higher precedence over the + then the = comes last.

Same goes with (3) and (4).

Can someone explain to me the reason?

Thanks!

Zagato

 

[mbaranski]

You're right, ++ has a higher precedence, but a++ and ++a mean totally different things:

a++ means evaluate a, and then increment it, so,
a++ + 4
if a = 3
is going to be 7,
and after the statement a will be 4.
This is because a is evaluated before it is incremented.

++a means increment a first, then evaluate it.
++a + 4
if a = 3
is going to be 8
and after the statement a will be 4.
This is because a is evaluated after it is incremented.

In the first example, a is considered a postfix operator because it comes after the value. In the 2nd example, it is a prefix operator. Ask again if you don't understand this. Also, this is c++ (I believe) not c, but no problem.

MWB As always, I hope that helped!

Disclaimer:
Beware: Studies have shown that research causes cancer in lab rats.

 

[dakuneko]

Yeah, it is C++.

Your post helped me understand better. Thank you!

I understand what you mean about the postfix and prefix operators. But I partly, I am still confused. In the statement

        a = b++ + 4;

if I follow the rule of precedence below

HIGH PRECEDENCE
Symbol Name or Meaning Associativity
++ Post-increment Left to right
-- Post-decrement
++ Pre-increment Right to left
-- Pre-decrement
- Unary minus
+ Unary plus
+ Add Left to right
- Subtract
= Assignment Right to left
LOWEST PRECEDENCE

it seems that I would have to evaluate ++ (Post-increment) first then the + (Add) and finally the = (Assignment).

If I would follow the rule above, it seems that the value that I will get for a if b= 1 would be 6

(b is evaluated then incremented then added to 4 which would follow the precedence ++, + then =)

but if this was actually run and according to what you said (which is correct) the end result for a would be 5.

(b is evaluated then added to 4 then incremented which would be +, =, ++)

That is why I am confused about this.

Could the reason behind this is because of how the statement gets stored in memory and partly because of the way it gets parsed?

Obviously, I am confused about how it has been evaluated. It seemed to me that it violated the rule of precedence. But that's how it is. Maybe this is a special case.

So, maybe if you have time, could you explain to me why it has violated the precedence rule for this case?

Thanks again!

Zagato

 

[dakuneko]

I've got another one. Consider the code below

	/* function foo */
	int foo(int *ey)
	{
		*ey = 1;
		return 1;
	}

	/* main */
	int main(void)
	{

		int a = 0;	

		/* 1 */	  printf("Return value of ptrf is %d. The value of a \ 
		                  in this statement is %d\n", foo(&a), a);
		/* 2 */	  printf("The value of a is now %d\n", a);
		return 0;
	}

The statement /* 1 */ has evaluated the assignment of a for the printf before using the value assigned to a by the function foo.

The printf of /* 1 */ should produce a value 1 for the return and still 0 for a. But when it reaches /* 2 */, the value of a would be 1.

Somebody help! Where in the rule of precedence does this come from?

Thanks!

Zagato

 

[mbaranski]

No, in hte stmt
a = b++ + 4;

the order is:

Get the value for b, increment b, add *the value for b* and 4, assign it to a.

Conversly,

a = ++b + 4

is:

Increment b, get the value for b, add *the value for b* and 4, assign it to a

So, the precdence for ++ is higher than addition, but it just depends on when you lookup the value for b, one does it before the ++ and one after the ++. This is why you see the different answers.

MWB. As always, I hope that helped!

Disclaimer:
Beware: Studies have shown that research causes cancer in lab rats.