Not able to find the square of the floating number ...

mybm1

Quote:

#include
#include

int main(int argc,int *argv[])
{
srand(12345); // generate same random number anytime it run
//srand((unsigned int)time NULL)
int i,j;
float a=0.2;
double x;
double sqrt(double);
printf("========================================================\n");
printf("Generated random number\t || Squaring of random number:\n");
printf("=========================================================\n");
for(i=0;i<100;i++)
{
double f= ((float)rand()/(float)(RAND_MAX))*a; //rand return integer value from 0 to RAND_MAX(system dependent)
  printf("%f\\t",f);
// printf("%f\t",((float)rand()/(float)(RAND_MAX))*a); // rand return integer value from 0 to RAND_MAX(system dependent)
printf("\t\t%f\n",x*f);
// printf("\t\t%f\n",f*f);
// printf("\t\t%f\n",(((float)rand()/(float)(RAND_MAX))*a) * ((float)rand()/(float)(RAND_MAX) *a));
}
  	FILE \*fout;
  	//FILE \*square;
  	fout=fopen("R\_Number","w");  // output file in write mode
  	j=0;
  	
  	srand(12345);                // to giving value 12345 generate same random number wen anytime it run
  	
  	while(j<100)                 // condition
  	{ 
  		 double f= ((float)rand()/(float)(RAND\_MAX))\*a;
  			 fprintf(fout,"%f\\t\\t\\t%f\\n",f,f\*f);
  			 
  	  //	 fprintf(fout,"%f\\n",((float)rand()/(float)(RAND\_MAX))\*a); // print file in output mode
  		j++;
  	}
  	fclose(fout);  // close the file
return 0;
}

double sqrt (double f)
{
double x, z, tempf;
unsigned long *tfptr = ((unsigned long *)&tempf) + 1;
tempf = f;
*tfptr = (0xbfcdd90a - *tfptr)>>1;
x = tempf;
z = f*0.5;
x = (1.5*x) - (x*x)*(x*z); //The more you make replicates of this statement
//the higher the accuracy, here only 2 replicates are used
x = (1.5*x) - (x*x)*(x*z);
return x*f;
}

Jochen Arndt

Where did you find that obscure sqrt() function? I guess it was originally using float rather than double. Remove that and use the C standard library sqrt() function by including math.h.

mybm1

actually i changed it to double n include math.h also but output seem same as before. source for square function

CPallini

Quote:

double sqrt (double f) { double x, z, tempf; unsigned long *tfptr = ((unsigned long *)&tempf) + 1; tempf = f; *tfptr = (0xbfcdd90a - *tfptr)>>1; x = tempf; z = f*0.5; x = (1.5*x) - (x*x)*(x*z); //The more you make replicates of this statement //the higher the accuracy, here only 2 replicates are used x = (1.5*x) - (x*x)*(x*z); return x*f; }

That is really messy! You are using uninitialized variables (that is garbage). It looks you want to implement the Babylonian method[^] but you do nothing for computing the initial guess. The following code is based on that very Wikipedia page.

void initial_guess(double r, int *pa, int *pn)
{
*pn = 0;

while ( r < 1.0)
{
r *= 100.0;
--(*pn);
}
while ( r >= 100.0 )
{
r /= 100.0;
++(*pn);
}

*pa = (r < 10.0) ? 2 : 6;
}

double square_root(double r)
{
int a, i, n;
initial_guess(r, &a, &n);
double x=1.0;

while (n < 0)
{
x/=10.0;
++n;
}
while(n > 0)
{
x *= 10.0;
--n;
}

x *= a;

for (i = 0; i<10; ++i) // 10 is the arbitrary number of iterations I chose
{
x = 0.5 * (x + r/x);
}

return x;
}

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

Jochen Arndt

You must remove your version of the sqrt() function. Otherwise it will be used instead of the standard library version. Your version uses ugly programming that directly accesses a single precision (float) value. By changing the type to double it won't work any more. [EDIT]: Forget that. It is an approximation that might be from http://www.azillionmonkeys.com/qed/sqroot.html[^]. But such hardware dependant code should not be used.

mybm1

thanks....

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

mybm1

using

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 squared number is:=%f\t",sum);

gives output but not the exact answer.. any solution actually i want to find the summation of that squared float number..

CPallini

How do you check the result?

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

mybm1

Quote:

Generated random number || Squaring of random number: ========================================================= 0.035679 ----- 0.001273 0.079935 ----- 0.006390 0.033320 ----- 0.001110 0.042424 ----- 0.001800 0.012387 ----- 0.000153 summation of squared number is:=0.010346

where the actual summation should be 0.10726

CPallini

You are wrong, the output of the following program:

int main()
{
int i;
double sum = 0.0;
double a[] =
{
0.035679,
0.079935,
0.03332,
0.042424,
0.012387
};

for (i=0; i<sizeof(a)/sizeof(a[0]); ++i)
{
double sq = square_root(a[i]);
printf("a[%d] = %g, sqrt(a[%d])=%g\n", i, a[i], i, sq);
sum += sq;
}
printf("sum of square roots = %g\n", sum);
}

is

a[0] = 0.035679, sqrt(a[0])=0.188889
a[1] = 0.079935, sqrt(a[1])=0.282728
a[2] = 0.03332, sqrt(a[2])=0.182538
a[3] = 0.042424, sqrt(a[3])=0.205971
a[4] = 0.012387, sqrt(a[4])=0.111297
sum of square roots = 0.971422

That is correct (at least according to Excel :-) ).

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

jeron1

Are you needing the sqaure or the square root? PS and the sum of your numbers is 0.010726.

"the debugger doesn't tell me anything because this code compiles just fine" - random QA comment "Facebook is where you tell lies to your friends. Twitter is where you tell the truth to strangers." - chriselst

mybm1

this is different program its squaring of number not square root...

mybm1

Actually i need square but its solve i got it but i am not able to get the summation value of the square number correctly .plz help me out..

CPallini

How do you compute the squares? This program:

#include <stdio.h>
int main()
{
double a [] =
{
0.035679,
0.079935,
0.033320,
0.042424,
0.012387
};
double sum, square;
int n;

sum = 0.0;
for (n=0; n<sizeof(a)/sizeof(a[0]); ++n)
{
square = a[n]*a[n];
sum += square;
printf("%g ----> %g\n", a[n], square);
}

printf("sum of squares: %g\n", sum);

return 0;
}

gives:

0.035679 ----> 0.00127299
0.079935 ----> 0.0063896
0.03332 ----> 0.00111022
0.042424 ----> 0.0017998
0.012387 ----> 0.000153438
sum of squares: 0.0107261

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