#include // hand trace practice int main(void) { // some have initial values int num1 = 8, num2 = 6; int sum, result; int trash = -4; // some crazy calculations result = 4 * num1 - 26 / num2 % 3; sum = num1++ + --num2; trash -= sum + result; // display results printf("num1 %d num2 %d sum %d result %d trash %d", num1, num2, sum, result, trash); printf(" "); return 0; } What are sum and result?

Question

#include <stdio.h>

// hand trace practice
int main(void)
{
   // some have initial values
   int num1 = 8, num2 = 6;
   int sum, result;
   int trash = -4;

// some crazy calculations
   result = 4 * num1 - 26 / num2 % 3;
   sum = num1++ + --num2;
   trash -= sum + result;

// display results
   printf("num1 %d   num2 %d   sum %d   result %d   trash %d", 
           num1, num2, sum, result, trash);
   printf("

");

return 0;
}

What are sum and result?

anonymous · Answer

compile and run it :-D http://codepad.org/ujqqgJTp

anonymous · Answer

ooh hand trace practice. so you weren't supposed to do what I just did :-P

lets initialize the stuff
$$num1 = 8$$
$$num2 = 6$$
$$trash = -4$$
this one gets a little tricky, but if you are absolutely unsure, refer to the white book pages 52 to 53. And lo! *, /, and % all have the same precedence, and are left to right associative. *, %, and / have higher precedence that subtraction.
I translated it to parens in the following expression:
$$result = (4*num1) - ((26/num2)mod3)$$
$$result = 32 - 1 = 31$$
the prefix -- decrement operator decreases the value of num2 before the rest of the expression is calculated.
$$num2 = 5$$
num1 uses the same old value of num 1, which is 8
$$sum = 8 + 5 =  13$$
the postfix ++ increment operator increments num1 after sum is assigned a value.
$$num1 = 9$$
the next one is straightforward
$$trash = trash - (13 + 31) = -4 -44 = -48$$

Therefore, you should get something like
$$num1 = 9, num2 = 5, sum = 13, result = 31, trash = -48$$