increment and decrement problem in c

[suman das]

slove this problem

include<stdio.h>
include<conio.h>
void main()
{
int i=1;
printf("%d",i++*++i);
getch();
}
if the ans is 3 explain it how;
there is a problem
it is found that the order of precedence of increment and decrement operator
with respect to other arithmetic operator is not well define.
in some computer the result may vary , it may be 3 and 4 both if we use torbo c++ for compilation. same for printing
printf("%d",i++*++i*++i)
result is expected 18 by RL rule by in my torbo c it gives 12 ans
plz write if u have any rule to sort it out.

[ilikepython]

slove this problem

include<stdio.h>
include<conio.h>
void main()
{
int i=1;
printf("%d",i++*++i);
getch();
}
if the ans is 3 explain it how;
there is a problem
it is found that the order of precedence of increment and decrement operator
with respect to other arithmetic operator is not well define.
in some computer the result may vary , it may be 3 and 4 both if we use torbo c++ for compilation. same for printing
printf("%d",i++*++i*++i)
result is expected 18 by RL rule by in my torbo c it gives 12 ans
plz write if u have any rule to sort it out.

The behaivior is undefined whenever you increment a variable more than once. The result can be anything and shouldn't be relied on.

[debojyoti]

++ operator has higher priority than *...

so in case pintf("%d",i++ * ++i);
the value of i is incremented by 1 in i++ but it is not assigned in i

in case of ++i value of i is incremented by i and it is assigned in i

the initial value of i is 1

try to dry run (i++ * ++i)....in i++ the value of i is incremented...as the value is not assigned the result of i++ operation is 1

but the value which in i becomes 2...after ++i operation the value of i becomes 3

multiply 1 and 3 the result is 3

apply the same logic in the later problem...hope u will get the correct result

[donbock]

Look here and here and here and here.