INTERVIEW QUESTIONS PROGRAMMING LANGUAGES C DETAILS

Question: main()
{
int i =10, j = 20;
clrscr();
printf("%d, %d, ", j-- , --i);
printf("%d, %d ", j++ , ++i);
}

Answer: A) 20, 9, 19, 10
in first -- printf "j" will be printed first then subtracted one from it. however from "i" one will subtracted before printing it on the screen

A) 20, 9, 19, 10
in first -- printf "j" will be printed first then subtracted one from it. however from "i" one will subtracted before printing it on the screen


Source: CoolInterview.com

In printf() evaluation happens from right to left. <--- . so --i is evaluated first and then j--.

When --i is evaluated, first it should decrement and then assign to i. So the value becomes 9.
When j-- is evaluated, first assign and then decrement. So j-- value will be 20 only at that time later it will be 19.

At this point of time i=9 and j=19.

Now the second printf(), Here also right to left. So ++i value will be(increment and then assign) 10. Where as j++(assign first then increment) its value will be still 19. After this printf() if we print the i and j values it will be 10 and 20.

Extra Information:
Whenever parameters are passed to a function, after evaluation they are stored in stack. similarly, --i value ie 9 is pushed into stack and later j-- value ie 20. So while printing on the screen, the first value ie j-- is poped first and later --i value. Hence we get 20,9 as the output of the first printf(). similarly we get 19,10 Source: CoolInterview.com

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