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Confusion with ++ operator

 
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Top Forums UNIX for Dummies Questions & Answers Confusion with ++ operator
# 1  
Old 08-23-2014
Code Confusion with ++ operator
Can anyone guide me whats happening in this program given below. I got the Output 7 7 12 49... i was expecting 5 16 9 25.
First is simple (3+1)*(3+1)
Second is again 3*3; i =4 now
Third i =5 then 5*5;
i don't know where i am going wrong!
Code:
#include<stdio.h>
#define PRODUCT(x) (x*x)
int main()
{
	int i=3,j,k,l;
	j=PRODUCT(i+1);
	k=PRODUCT(i++);
	l=PRODUCT(++i);
	printf("%d %d %d %d\n",i,j,k,l);	
	return 0;
	
}

# 2  
Old 08-23-2014
Quote:
Originally Posted by Abhishek_kumar
Can anyone guide me whats happening in this program given below. I got the Output 7 7 12 49... i was expecting 5 16 9 25.
First is simple (3+1)*(3+1)
Second is again 3*3; i =4 now
Third i =5 then 5*5;
i don't know where i am going wrong!
Code:
#include<stdio.h>
#define PRODUCT(x) (x*x)
int main()
{
	int i=3,j,k,l;
	j=PRODUCT(i+1);
	k=PRODUCT(i++);
	l=PRODUCT(++i);
	printf("%d %d %d %d\n",i,j,k,l);	
	return 0;
	
}

There is a big difference between a macro expansion and a function call. Your assignments to j, k, and l expand to:
Code:
	j=i+1*i+1; /* 3 + 3 + 1 -> 7 */
	k=i++*i++; /* 3 * 4 -> 12 leaving i set to 5 */
	l=++i*++i; /* 6 * 7 -> 42 leaving i set to 7 */

Are you sure you got 49 instead of 42 as the last number printed?
# 3  
Old 08-23-2014
I get 7 7 9 49

Apparently using incremental ops multiple times in the same statement is undefined in C.
# 4  
Old 08-23-2014
Quote:
Originally Posted by neutronscott
I get 7 7 9 49

Apparently using incremental ops multiple times in the same statement is undefined in C.
Yes, the behavior in those cases is unspecified (except that i should be 7 at the end). But with compilers I've used in the past, the results I quoted was what you would usually get. I had seen compilers that would give one of:
Code:
7 7 9 36
7 7 9 49
7 7 16 36
7 7 16 49

before (both with or without increments in all cases), but I had never seen one that did:
Code:
7 7 12 49

with immediate increments on i++ and using both values (3 * 4) on one, but doing both ++i operations and using the final value (7 * 7) on the other. Oh, well.
# 5  
Old 08-23-2014
My understanding is that * and ++expr operators have the same precedence in the C language but that ++exprs are always evaluated before any *.

Thus
Code:
i=5
(++i) * (++i)

gives 49 with both gcc and g++
# 6  
Old 08-24-2014
Quote:
Originally Posted by fpmurphy
My understanding is that * and ++expr operators have the same precedence in the C language but that ++exprs are always evaluated before any *.

Thus
Code:
i=5
(++i) * (++i)

gives 49 with both gcc and g++
My understanding is that the actual increment of i can occur at any time before the next sequence point. In the statements:
Code:
	k=i++ * i++;
	l= ++i * ++i;

the ; is a sequence point; the * is not a sequence point.

The submitter hasn't specified which compiler is being used in this example. Using the compiler on OS X 10.9.4 to build and run the sample program provided in the 1st message in this thread yields:
Code:
$ make t
cc     t.c   -o t
t.c:7:13: warning: multiple unsequenced modifications to 'i' [-Wunsequenced]
        k=PRODUCT(i++);
                   ^~
t.c:2:21: note: expanded from macro 'PRODUCT'
#define PRODUCT(x) (x*x)
                    ^
t.c:8:12: warning: multiple unsequenced modifications to 'i' [-Wunsequenced]
        l=PRODUCT(++i);
                  ^~
t.c:2:21: note: expanded from macro 'PRODUCT'
#define PRODUCT(x) (x*x)
                    ^
2 warnings generated.
$ ./t
7 7 12 42

It is perfectly legal for a compiler to perform the increments before the multiplication, but the C Standard doesn't require it to behave that way. And, if the compiler the submitter uses did behave that way, the output would be:
Code:
7 7 9 49
    or
7 7 16 49

rather than:
Code:
7 7 12 49

 
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