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why not break;

 
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why not break;
kornni
7 Jul, 2008 - 09:16 AM
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why this program do not break; when u==0

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

#define pi   4 * atan(1.0)

float t;
double u;
int main()
{
    u = 100;
    for(t = 0; t <0.2; t += 0.01)
    {    
          u = u*sin(0.5*pi) - u*sin(0.6*pi) + u*sin(0.7*pi);
          printf("%f\n",u);
          
          if(u==0)
          {
              break;
          }      
    }
}


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lordms12
RE: Why Not Break;
7 Jul, 2008 - 10:25 AM
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While dealing with double or float it is not a good way to compare with equal because of accuracy problems, somehow I mean that .00009 might be zero for you (according to accuracy you want) but is not zero for computer
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KYA
RE: Why Not Break;
7 Jul, 2008 - 10:53 AM
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Works for me if u ever equals zero. Use a test value to ensure the mechanics are correct--seems fine.
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captainhampton
RE: Why Not Break;
7 Jul, 2008 - 11:07 AM
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The operations you are performing are too precise to get a rounded zero value, as opposed to the decimal values you are getting from sin, cos, etc. Perhaps try a less than condition except == 0
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born2c0de
RE: Why Not Break;
8 Jul, 2008 - 10:47 PM
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You are comparing a double with an integer, and conversion may result in a loss of precision.

Use this instead:
cpp
if( u == 0.0 )

This way the compiler interprets 0.0 as a double and not an integer. (if 0.0f is mentioned, the compiler will consider it as a float)
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Tom9729
RE: Why Not Break;
9 Jul, 2008 - 05:57 PM
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I haven't really looked at your algorithm but keep in mind you can have things like negative zero as a floating point value, which will not equal normal zero.
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