Explain the output of these C codes

general
1 
#include <stdio.h>
int main ()
{
	int i = 4;
	int x = ++i + ++i + ++i;
	printf ("%d", x);
	return 0;
}

Output : 19

#include <stdio.h>
int main ()
{
	int i = 0;
	while (i++ <= 10);
	printf ("%d", i);
	return 0;
}

Output : 12

2

I might be wrong(as i am applying reverse engineering) but still-
I think for first the compiler will use right precedence.
So the expression will be evaluated as ++i + (++i + ++i)
Since ++ has higher precedence so it will be ++i +(++i + 5)
then ++i +(6+5)
now maybe this addition will be with respect to memory location of i
since value of both memory locations(i) is changed to 6 instead of 6+5 ,6+6 is added and the result is 12
this 12 will be stored in a free memory location(perhaps pre-allocated by the program but not alive here)
then 7+12 is added and answer is 19.

2)in loop where i is 10 10 is compared but i is incremented to 11.So the condition fails next time.But while evaluating the condition its value changes to 12 which is printed.

2 Likes
3

The second explanation seems correct. Thanks.

4

note semicolon after while loop in 2nd Question ,and for the first question the answer does vary according to the compiler,it is either 18(5+6+7) or 19(6+6+7) FOR THE reasons as explained earlier

5

what’s wrong with the semi-colon?