Reading the program/output question

**benrogers** · 01-26-2011

Code:

int main()
{
   int x = 5;
   double d = 3.5;
   double y = 2;

while (x > 3)
  {
      printf(“%d\n”, x);
      d += (x % 2);
      x = x – 1;
 }

/* a */

   x = d > d ? x : (d / x);

   if (x < 1)
     y = x + 3;
   else if (x == 1)
     y = y – x;

   if (x == 1)
     y = y * y;


/* b */
return 0;
}

Please explain why:

a) what is the value of x at the spot marked a? - I see it as 4. why is it really 3?
b) what is the output of the program ?
5
4 - why 4?
c) what is the value of d at the spot marked a? - 4 because it's d +=?
d) what is the value of x at the spot marked b? why 1
e) what is the value of y at the spot marked b? why 1

**tabstop** · 01-26-2011

a) Because 4 is bigger than 3. You can't stop the loop if the loop condition is still true.

c) The answer can't possibly be 4 -- you've got a half-integer to start with (3.5), you're adding integers to it, therefore you will always have a half-integer value for d.

**benrogers** · 01-26-2011

c) 4.5 is the answer.

**anduril462** · 01-26-2011

Your best bet for something like this is to make a table with one column for each variable, and one row per instruction:

Code:

 x |  d  |  y
---+-----+-----
 5 | 3.5 |  2      <-- Initial values, start of program
 5 | 3.5 |  2      <-- x > 3, enter loop
 5 | 3.5 |  2      <-- program prints x (5)
 5 | 4.5 |  2      <-- x % 2 = 5 % 2 = 1, d += 1
...

It's tedious, but a great way to figure out what the computer is doing when it runs your code.

**~~CommonTater~~** · 01-26-2011

Code:

   x = d > d ? x : (d / x);

Under what possible condition can d be greater than d?

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