can anyone help how to find summation of float number?

[mybm1]

for(i=0;i<5;i++)
{
float f= ((float)rand()/(float)(RAND_MAX))*a; //rand return integer value from 0 to RAND_MAX(system dependent)
//float y[5];

        printf("%f\\t",f);


        y=square(f);
        printf("\\t\\t%f\\n",y);
        
           }




    for(i=0;i<5;i++)
                 {  //float y,sum;
                float f= ((float)rand()/(float)(RAND\_MAX))\*a;
                y=square(f);
                    sum +=y;
             }
    printf("\\n Summation Of Square Number is:=%f\\t\\n",sum);

i have print at the 1st for loop...:(

[enhzflep]

Except for a rather small subset of numbers, calculations with computers involving floating-point numbers are approximate. Integer arithmetic doesn't suffer from this pitfall. However, if you are talking about the result of the code I posted - then consider reviewing the printf statements - in each of them I have used the %.2f format specifier to tell printf to only print 2 digits after the decimal place. The point is - the answer is as close to exact as floats will give you, it is only the display that is (grossly) approximate. :) You can get a less approximate result printed if you change the print specifiers to %f

[CPallini]

#include

#define N 5

double summation(double g[], int size);

int main()
{
double a[N] =
{
0.035679,
0.079935,
0.033320,
0.042424,
0.012387
};

double s[N];

int i;
for (i=0; i
THESE PEOPLE REALLY BOTHER ME!! How can they know what you should do without knowing what you want done?!?!
-- C++ FQA Lite

[CPallini]

As far as I can understand, he is trying to imlpement the "Kahan summation algorithm"[^].

THESE PEOPLE REALLY BOTHER ME!! How can they know what you should do without knowing what you want done?!?! -- C++ FQA Lite

[mybm1]

CPallini i have given just for 5 number as demo but in real my count is above 60000 ?and i am searching for the summation for all the 60000+ value which give correct result.it wouldt be possible for what?

[enhzflep]

Seems like a reasonable assumption to me. I'd seen your mention of it earlier, but hadn't seen a confirmation of this so guessed (wrongly?) where the problem lay. Thanks for the impetus to re-investigate the method - I seem to remember using it for something a while back, but brain-rot defeats me.. :sigh:

[mybm1]

can u please kindly explain your assumption?becoz een though i put %.2f its doesnot seem much change in the answer that y..

[CPallini]

You cannot modify my program for processing more items, can you? :rolleyes:

THESE PEOPLE REALLY BOTHER ME!! How can they know what you should do without knowing what you want done?!?! -- C++ FQA Lite

[enhzflep]

Certainly. Both your code and the pseudo-code in the wiki page that CPallini linked to appear quite similar. This, combined with your mention of the result only being an approximation seems to further support your need for a more accurate result of summing. (which the method may facilitate)

[Lost User]

And in the second loop you will have a different set of numbers, as returned by the calls to rand(). I also cannot find a reference to square, is this another function in your code?

[mybm1]

:-D thank you finally i got it after full torture to u and others too..

[mybm1]

omg u were right ,finally i got the problem was only with the where i gave the print statement..

[mybm1]

ya it is thank fren i got it...matter was with the printf statement i gav at second loop ,wen i pass the printf statement inside the 1st loop itself n removing the second for loop as condition where same ...den i got the exact answer..

[CPallini]

You are welcome.

THESE PEOPLE REALLY BOTHER ME!! How can they know what you should do without knowing what you want done?!?! -- C++ FQA Lite

[Satya Chamakuri]

hi, I don't know what's the use of summation in your code. If you want summation of float numbers generated numbers only means, the following piece of code is enough i guess.. See once

srand(12345);
float z=0.0;
float a=0.2,y=0.0;
int i;
for(i=0;i<5;i++)
{
float b=((float)rand()/(float)(RAND_MAX))*a;
printf("%f\n",b);

    z=z+b;
}
printf("summation:\\n%f\\n",z);