What is the order of evaluation of the statements in this code?

Question

I came across this question.

#include <stdio.h>
int main()
{
int k=8;
int x=0==1||k++;
printf("%d  %d",x,k);
    return 0;
}

The output is 1 9.

As this answer suggests that

Unlike the bitwise | operator, the || operator guarantees left-to-right evaluation; there is a sequence point after the evaluation of the first operand. If the first operand compares unequal to 0, the second operand is not evaluated.

I am unable to understand how the statement int x=0==1||k++ is evaluated,due to which the value of x and k becomes 1,9 respectively.

Can someone explain how such statements are evaluated by the compiler in c ?

It can be written more unambiguously as `int x = ( (0 == 1) || (k++) );` - does that make it clearer? Which part do you not understand? — Oliver Charlesworth, Jul 17 '16 at 13:01
@ OliverCharlesworth why it can't be written as `int x =(0==(1 || (k++))) ` ? — asad_hussain, Jul 17 '16 at 13:05
There are exactly two statements in this code (one expression-statements and one return-statement). They are evaluated top to bottom: first `printf("%d %d",x,k);` then `return 0;`. The two other lines are declarations, not statements. — n. m. could be an AI, Jul 17 '16 at 13:06

eRaisedToX · Accepted Answer · 2016-07-17T13:31:11.923

5

"Unlike the bitwise | operator, the || operator guarantees left-to-right evaluation; there is a sequence point after the evaluation of the first operand. If the first operand compares unequal to 0, the second operand is not evaluated."

yes it's true ...

to make it clear first make sure that you are aware of these basics

1) (1 || any_var) is 1

2) Operator precedence is as follows

++

then

==

then

||

NOW coming to your doubt of || vs |

note that | (single pipe) operator ..,will execute both LHS and RHS , no matter what)

whereas

|| (double pipe) evaluates LHS of || first and if it is 1 it need not evaluate RHS (for speed)

RHS of || operator will not be evaluted if LHS comes out to be true.

but here 0==1 is false i.e 0==1 returns 0

hence RHS will be evalauted

so the statement

k++ is executed

but 8 is used (as property of post increment operator says--> first use then increment)

so 0||8 is definitely true (1) so x evaluates to 1

and then k is incremented after sequence point ie k is made equal to 9

hence output x=1 and k=9

I hope it clears your doubt :)

edited Jul 17 '16 at 13:31

answered Jul 17 '16 at 13:02

eRaisedToX

3,221
2
22
28

I still don't get you. Why is 0||8 true? – VatsalSura Jul 17 '16 at 13:09
@VatsalSura - What else would it be? – Oliver Charlesworth Jul 17 '16 at 13:11
@VatsalSura Because 8 is not zero, hence `||` treats it as "true", and returns 1 for the result. – Sergey Kalinichenko Jul 17 '16 at 13:11
@vatsal by true I mean 1 ...,yeah ( ! 0 || x) is always 1 , anything other than 0 when ORed with anything turns out to be 1 – eRaisedToX Jul 17 '16 at 13:12
So, why the value of `x` is `1` and not `8` or `9`? – VatsalSura Jul 17 '16 at 13:16
@ vatsal why do you think it should be 8 .. , forget about 9 for now , k is not yet incremented so it wont be 9 at all .., NOW as I already commented that as per OR (||) oprerator , anything other than 0 when ORed with anything returns 1 – eRaisedToX Jul 17 '16 at 13:19
1

@VatsalSura the `||` operator always gives either 0 or 1 – M.M Jul 17 '16 at 13:25
Ok, Thank you guys, now I understood it. – VatsalSura Jul 17 '16 at 13:26

score 1 · Answer 2 · answered Jul 17 '16 at 13:23

The Operator || (OR) evaluates to true in the cases:

ex: A || B

A is true,
B is true,
Both are true

Because this operation uses Short-Circuit Evaluation if A is evaluated to true, it means that the statement is true already and it won't evaluate B.

In your case 0==1 (0 equals 1) is clearly false, so it will evaluate k++. k++ is a tricky one (in my opinion). In the world of C true/false evaluation is based on being 0 or not (except for the times false means less than 0...) but in true/false evaluation 0 is false, everything else is true.

K++ means evaluate K then increment, so, if K is 0 it will be false and become 1, if it is anything else, it will be true and then increment.

In your case k == 8 so the result of k++ is TRUE and k will become 9. K being true means x evaluation resulted in TRUE (it was FALSE OR TRUE).

So the output is 1(True) 9(8++)

babon · Answer 3 · 2016-07-17T13:32:59.413

x is 1 because the expression: 0==1||k++ turns out to be true (which is 1 in C land). Why you ask? There are two sequence points here: 0 == 1 and k++. Since the first sequence point evaluates to false (0 in C land), the second sequence point is evaluated (because the short circuit operator is ||). The second sequence returns true (or 1). So, you the entire expression breaks down to: 0 || 1. Hence x is 1.

k is 9 because of k++;

HTH.

What is the order of evaluation of the statements in this code?

3 Answers3