incorrect result -_-

[Tubs]

code:

#include <stdio.h> #include <math.h> 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.

[md]

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:

Pascal:

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<float> or static_cast<double> (I think that's right... you can always use (double)... ).

c++:

double foo_cpp(int x, int y) {     return x / static_cast<double>y; }

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

[Hikaru79]

For example:

code:

#include <stdio.h> #include <math.h> 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.

[Tubs]

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

[md]

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

[Hikaru79]

Cornflake wrote:

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

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

[wtd]

Cornflake wrote:

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

[syntax="cpp"]static_cast<double>(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]

[md]

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