OOPS - could any one explan me what happens in this program. main(){ int i=10; i=(++i)/(i++)

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could any one explan me what happens in this program.

main(){
int i=10;
i=(++i)/(i++);
printf("i=%d",i);
}

Total Answers and Comments: 11

Last Update: March 09, 2008 Asked by: vaasu

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May 25, 2006 00:39:21

Umme Zakira

RE: could any one explan me what happens in this progr...

the value of i in the given program is 2

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May 25, 2006 06:34:06

rajan

RE: could any one explan me what happens in this progr...

In c it will do some right to left approach so the value of i 1.2

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May 25, 2006 13:56:04

srinivas

RE: could any one explan me what happens in this progr...

i know that the output is 2.but i want how the complier gives it.

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May 29, 2006 05:55:13

Malathi

RE: could any one explan me what happens in this progr...

Here ++i is previx expression -- i will be incrementd before the statement and i++ is postfix --> I will be incremented after the statement .

i (++i)/(i++); --- 1. ++i --> i will be incremented first.-> 11

2. 11/i 11/10 1.1

3. (i++) 1.1++ 2.1

printf( i d i); --> d id for integers therefore 2.1 is trucated to 2.

Bye

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May 31, 2006 03:18:52

RE: could any one explan me what happens in this progr...

Reason is quiet simple

i (++i)/(i++); it will resolve

i 11/10 which give 1 being int

now post increment works and increment new val of i thus giving 2.

Thanks

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June 21, 2006 15:06:58

amit

RE: could any one explan me what happens in this progr...

The output of this program will be undefined. In C and C++ if you read a variable twice in an expression where you also write it the result is undefined.for further details : http://www.research.att.com/~bs/bs_faq2.html#evaluation-order

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June 28, 2006 07:51:54

swapna s

RE: could any one explan me what happens in this progr...

Solution:-

i 10;

i (++i)/(i++)

++i is preincrement the value of i before executing the statement.

i++ is post increment the value of i after executing the statement

Here ++i 11

i (++i) / (i++); // 11/11 1

After executing the above statement post increment operator (ie i++)

is excuted. Becaz of that i's value is incremented to 2.

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July 03, 2006 01:41:06

sami ahmed

RE: could any one explan me what happens in this progr...

i agree with swapna s

but the output will b 1..

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January 26, 2008 18:24:43

zdmytriv

Member Since: January 2008 Contribution: 8

RE: could any one explan me what happens in this program. main(){ int i=10; i=(++i)/(i++); printf("i=%d",i); }

swapna was right

output: 2

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February 26, 2008 01:13:14

#10

chethan

Member Since: November 2005 Contribution: 3

RE: could any one explan me what happens in this program. main(){ int i=10; i=(++i)/(i++); printf("i=%d",i); }

1. i ++ will get execute so the value will be 10 . it wont increment.
2. ++i will be incremented to 11
3. 11/10 will result to 1
4.i 1 will get execute.
5. keep in mind still increment is not done.. this is done after exectuing or ;
6. after executing i 1 ';' will be executed
7. i++ will get incremented to 2

so the answer is 2.

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