doubt

Question

#include <stdio.h>
 
int main(void) {
	static int a[10];
	int x=0;
	a[x]=++x;
	printf("%d %d %d",a[0],a[1],a[2]);
	return 0;
}

my doubt here is [ ] has high precedence over ++ which in turn has precedence over =, so i think it a[x] = ++x; is evaluated in 3 phase like,

phase1: a[x] -> (a+x)* -> (a+0)* -> a*

phase2: ++x -> x=1

phase3: a*=1 or a[0]=1

so finally answer should be 1 0 0, but it is giving 0 1 0 as output.

Comments

I think "assignment operator evaluates an expression of the RHS and assign the resultant value to a variable of the LHS"  this is the reason behind it.

Hence while executing phase 3 value of x will be also changed on LHS.

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That is an undefined behaviour.

see this: https://stackoverflow.com/questions/367633/what-are-all-the-common-undefined-behaviours-that-a-c-programmer-should-know-a

and lines just above "see also" in: https://en.wikipedia.org/wiki/Undefined_behavior

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UNDEFINED BEHAVIOR. Because of this line a[x] = ++x.

You can check this link for more.

https://stackoverflow.com/questions/4176328/undefined-behavior-and-sequence-points

https://gateoverflow.in/62411/undefined-behaviour-in-c

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@Rishabh, 

the link you gave is different qsn, there it is a[x]=x++ //postfix operator, it can give undefined behavior since [ ] and ++(postfix) has equal precedence, but here i have given prefix operator.

though i got the reason, this is because reading the value of the object for any other purpose than determining the value to be stored is also undefined behavior.\

thankyou!!

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thankyou @Hemant @Rishabh.