CODEWITHSUNDEEP
c
ixnisarg
ixnisarg

Reputation: 117

Why This Code Prints Second statment?

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");
            }
    }

Upvotes: 0

Views: 85

Answers (5)

2501
2501

Reputation: 25752

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.

Upvotes: 2

user5478656
user5478656

Reputation: 276

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.

Upvotes: 0

msc
msc

Reputation: 34598

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.

Upvotes: 0

mbvreddy
mbvreddy

Reputation: 15

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)

Upvotes: -3

babon
babon

Reputation: 3774

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

Upvotes: 3

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