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Tubs

Mon Oct 03, 2005 8:28 pm


incorrect result -_-

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#include 
#include 

int main(int argc, char *argv[])
{

double thing;

thing = pow((27/1),(1/3));

  printf ("%.2f", thing);
  system ("PAUSE") ;
  return 0;
}


This returns 1.00 where it should return 3.00. (It returns one no matter what input is in the power function so i assume the problem is there). Any suggestions? Using dev c++ compiler.


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md

Mon Oct 03, 2005 8:41 pm



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well... 27/1 is 27 and 1/3 is one third. The problem is that you're using integers. When you devide one integer (say 1) by another (say 3) the compiler does the division, but instead of returning a floating point fraction it returns an integer value (in this case it returns 1, the value it returns is documented somewhere...). In pascal this does the same as:

function foo_pascal(x:integer; y:integer):integer;
begin
    foo := x div y;
end;


To get the fractional value in C or C++ you must first cast one of the values into a float or a double. There are by far many ways to do it, but the one that wtd would probably insist upon is a static_cast or static_cast (I think that's right... you can always use (double)... ).


double foo_cpp(int x, int y)
{
    return x / static_casty;
}


As for power()... I'm fairly certain that there's nothing wrong with it, but see what happens when you use the casting.


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Hikaru79

Mon Oct 03, 2005 8:43 pm



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For example:#include 
#include 

int main(int argc, char argv[])
{

float thing;

thing = pow((27.00/1.00),(1.00/3.00));

  printf ("%f",thing);
  return 0;
}
This will return 3.00. I just converted all those ints to doubles manually.


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Tubs

Mon Oct 03, 2005 8:45 pm



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thanks... i am kinda creeped out that you go to UW though. im like 15 minutes away ;) WLU


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md

Mon Oct 03, 2005 8:45 pm



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Sure you can do it manually... but then what happens if you change the values for variables? :P


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Hikaru79

Mon Oct 03, 2005 9:16 pm



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Sure you can do it manually... but then what happens if you change the values for variables? :P
Oh, no, I agree that your way is the solution. I was just giving a solid example to demonstrate what you were saying. Obviously the function has to be able to evaluate other values besides the cube root of 27 :P


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wtd

Mon Oct 03, 2005 9:23 pm



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There are by far many ways to do it, but the one that wtd would probably insist upon is a static_cast or static_cast (I think that's right... you can always use (double)... ).

[syntax="cpp"]static_cast(int_value)[/code]

Would be correct.  However, the original poster is clearly using C, as evidenced by the use of stdiol, so instead we use:

[syntax="c"](double)int_value[/code]


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md

Mon Oct 03, 2005 10:25 pm



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thanks wtd, wasn't quite sure on the syntax there