Sine values in integer format

i am facing the problem of displaying the sine values in integer format. I am trying to create a sine look up table. Just for testing i showing 10 values. I am getting fractional part but want to convert to integer and store in an array, but not able to convert. what is the mistake

Code:

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

int main(int argc, char const *argv[])
{
    int index=0;
    printf("fractional values \n");
    for(index=0; index <= 10; index++)
    {
        printf("%f,\n", 1000.0*sinf(2*3.14159*index/1000));
    }
    printf("Converted to integer values \n");
    for(index=0; index <= 10; index++)
    {
        printf("%d\n",(int)1000*sinf(2*3.14159*index/1000) );
    }
    return 0;
}

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output is like this

Code:

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fractional values
0.000000,
6.283139,
12.566029,
18.848424,
25.130074,
31.410731,
37.690152,
43.968081,
50.244275,
56.518488,
62.790463,
Converted to integer value
0
270415556
-1928745506
1438052426
-2118917277
-1456629062
-447103537
354588366
1679953399
-753718746
-427015548

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Quote:

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Originally Posted by Satya

i am facing the problem of displaying the sine values in integer format. I am trying to create a sine look up table. Just for testing i showing 10 values. I am getting fractional part but want to convert to integer and store in an array, but not able to convert. what is the mistake

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You cast 1000 to a integer (but 1000 is allready a integer).
You include math.h, so why do you don't use the defined M_PI ?
The line should look like this:

Code:

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printf("%d\n",(int) (1000.0 * sinf(2 * M_PI * index / 1000.0)));

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Yes got it. Thanks.