Why This Code Prints Second statment?

Question

Why This code fail in first If statement?

My prediction getting wrong as per associations and precedence.

 #include<stdio.h>

    void main()
    {
            int i=10;
            if(i==i--)
            {
                printf("In 1:%d\n",i);
                printf("TRUE 1\n");
            }   
            i=10;
            if(i==--i)
            {
                printf("In 2:%d\n",i);
                printf("TRUE 2\n");
            }
    }

There's no sequencing in the expressions `i == i--` and `i == --i`, so you have *undefined behavior*. — Some programmer dude, Sep 30 '16 at 07:09
For the technical bits, please read about [evaluation order and sequencing](http://en.cppreference.com/w/c/language/eval_order). — Some programmer dude, Sep 30 '16 at 07:12

score 3 · Answer 1 · answered Sep 30 '16 at 07:11

3

i==i-- is undefined behaviour. Please check this: http://c-faq.com/expr/ieqiplusplus.html and this: http://c-faq.com/expr/seqpoints.html

answered Sep 30 '16 at 07:11

babon

3,615
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2501 · Answer 2 · 2016-09-30T07:28:17.483

The expression i==i-- will cause undefined behavior because there is no sequence point between the two evaluations of i and i--. This means that anything can happen and at that point the program no longer produces meaningful output.

The same is true for the expression i==--i

If an object is read and also modified without a sequence point separating the two events the behavior is undefined¹. In this case the same object is modified (side effect): i-- and read (value computation): i, without a sequence point.

Correct code would separate the two expressions with a sequence point (character ;):

const int i1 = i;
const int i2 = i--;
if( i1 == i2 )
{
    //...
}


const int i3 = i;
const int i4 = --i;
if( i3 == i4 )
{
    //...
}

¹ (Quoted from ISO/IEC 9899:201x 6.5 Expressions 2):
If a side effect on a scalar object is unsequenced relative to either a different side effect on the same scalar object or a value computation using the value of the same scalar object, the behavior is undefined.

user5478656 · Answer 3 · 2016-09-30T07:23:30.400

0

There would be no compilation error. It does not go inside the first if statement as it is undefined behavior (as user babon mentioned).

The behavior actually depends on the compiler you're using.

edited Sep 30 '16 at 07:23

answered Sep 30 '16 at 07:13

user5478656

266
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9

score 0 · Answer 4 · edited May 23 '17 at 12:06

0

Undefined behaviour because post and pre increment depend on compiler. Please see stack overflow question here.

C99 standard - 6.5 Expressions, §2

Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression. Furthermore, the prior value shall be read only to determine the value to be stored.

edited May 23 '17 at 12:06

Community

1
1

answered Sep 30 '16 at 07:20

msc

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score -3 · Answer 5 · answered Sep 30 '16 at 07:13

-3

I did not quite get your question. Are you asking "why does statements in only first if gets executed"? If yes, following is the reason

i-- evaluates the expression and then decrements --i decrements first and then evaluate expression. So, second condition evaluates to 'false' (10 == 9)

answered Sep 30 '16 at 07:13

mbvreddy

15
2

what about second if statement? @Bhaktvatsal – ixnisarg Sep 30 '16 at 07:15
2

@M.SChaudhari no ; decrement and evaluation occur with no sequencing defined. Which is why the behaviour is undefined. – M.M Sep 30 '16 at 07:31

Why This Code Prints Second statment?

5 Answers5