Hi there all,

Could someone please tell me how I could increase my precision for floating point arithematic?

My requirement is that I add a very small value of the order 10^-7 with a relatively big value, say 36.63 and then I want multiply it with 10^7. The problem Im facing using float nos alone is that when i sum up the numbers i get 36.63 itself and the former is not padded to the value 36.63.
My problem here make a significant difference in values because rigtht after the summing i multiply it with 10^7.

Could somebody tell me how I could go about this problem?

Thanks in advance

## Recommended Answers

use double :)

Try cout.precision(25);

I guess you did not understand the problem.

I am adding two numbers-1.37*10^-5+6.3, in C++ when we do floating point arithmetic it automatically truncates the sum. And this is where I have the problem, I don't want it to truncate at all. because after i get the sum, i multiply …

> My requirement is that I add a very small value of the order 10^-7 with a relatively big value, say 36.63
floats have about 6 decimal digits of precision, doubles about 15.
10^-7 is more than 6 digits away from 36.63 so your really small number is effectively zero.

## All 16 Replies

use double :)

I guess you did not understand the problem.

I am adding two numbers-1.37*10^-5+6.3, in C++ when we do floating point arithmetic it automatically truncates the sum. And this is where I have the problem, I don't want it to truncate at all. because after i get the sum, i multiply it with 10^7,so if it truncates there is a significant difference in values. Could somebody tell me how to go about this problem?

Thank you.

Try cout.precision(25);

I guess you did not understand the problem.

I am adding two numbers-1.37*10^-5+6.3, in C++ when we do floating point arithmetic it automatically truncates the sum. And this is where I have the problem, I don't want it to truncate at all. because after i get the sum, i multiply it with 10^7,so if it truncates there is a significant difference in values. Could somebody tell me how to go about this problem?

Thank you.

There's a problem with your compiler friend !
I just tried the sum and it doesn't truncate anything. (VS 2008)

Float has a range of 3.4E +/- 38 so it's way more than you need !

Maybe it you paste the code you use ...

> My requirement is that I add a very small value of the order 10^-7 with a relatively big value, say 36.63
floats have about 6 decimal digits of precision, doubles about 15.
10^-7 is more than 6 digits away from 36.63 so your really small number is effectively zero.

Using double will buy you some head room, but won't solve the underlying problem.

Two choices
- use a math library with arbitrary precision like GMP
- rearrange your expressions so that precision is preserved as long as possible.
Eg.

``````float small = { };
float big;
for ( i = 0 ; i < 10 ; i++ ) big += small[i];``````

would become

``````float small = { };
float big;
float smallish = 0;
for ( i = 0 ; i < 10 ; i++ ) smallish += small[i];
big += smallish;``````

Each small by itself is too small to affect big, but by combining them all together you end up with something which can affect it.

Hi Vale82 and Salem,
I will show my code to show whats happening.

``````Nx=1280;
Ny=960;
double pinhole_ccd_D_rec=6.3;
double pinDrec=pow(pinhole_ccd_D_rec,2);
double k=2*pi/lamda;[1.18105e+07]
for (m=0;m<Nx;m++)
{ for(n=0;n<Ny;n++)
{ 	x[n][m]=(1+m-Nx/2); y[n][m]=(1+n-Ny/2);
r[n][m]= pow((x[n][m]*dx),2)+ pow((y[n][m]*dy),2);
PD=k*sqrt(r[n][m]+pinDrec);
}
}``````

My ans to PD is more or less the same because of the truncation.
First value of r[n][m]-1.37*10^-5,pinDrec=36.96, I take the sum, take the sqrt and then multiply with a huge value k. So even the small changes should be reflected in the sum r[n][m]+pinDrec. I am using VC++ 6.0.

Could you guide me where I am going wrong

Hi chiraag,

I've tested with a somewhat simplified version of your example:

``````const int Nx = 1280;
const int Ny = 960;

double x;
double y;
double r;

double pinhole_ccd_D_rec = 6.3;
double pinDrec = pow(pinhole_ccd_D_rec,2);
double pi = 3.14, lamda = 1.18105e+07;

double k = 2 * pi /lamda;//[1.18105e+07]
double dx = 1, dy = 1;
double PD = 0;

for (int m = 0; m < Nx; m++)
{
for(int n = 0; n < Ny; n++)
{
x = (1 + m- Nx / 2);
y = (1 + n- Ny / 2);

r= pow((x * dx),2) + pow((y * dy), 2);

PD = k * sqrt(r + pinDrec);
}
}``````

The thing is that PD is quite different every time. Maybe my version of the code is missing something ?!

My advice would be to get a new compiler ( download the express VS 2008 for C++, it's free on the microsoft site ) and try your code in that one. VS 6.0 has a LOT of bugs and even if the 2008 version still isn't working for you it's still an upgrade you MUST make!

One more thing, when you're testing the values, I hope you're in debug and looking directly at them! If you're printing them out on the console ... all bets are off :)

Hi chiraag,

I've tested with a somewhat simplified version of your example:

``````const int Nx = 1280;
const int Ny = 960;

double x;
double y;
double r;

double pinhole_ccd_D_rec = 6.3;
double pinDrec = pow(pinhole_ccd_D_rec,2);
double pi = 3.14, lamda = 1.18105e+07;

double k = 2 * pi /lamda;//[1.18105e+07]
double dx = 1, dy = 1;
double PD = 0;

for (int m = 0; m < Nx; m++)
{
for(int n = 0; n < Ny; n++)
{
x = (1 + m- Nx / 2);
y = (1 + n- Ny / 2);

r= pow((x * dx),2) + pow((y * dy), 2);

PD = k * sqrt(r + pinDrec);
}
}``````

The thing is that PD is quite different every time. Maybe my version of the code is missing something ?!

My advice would be to get a new compiler ( download the express VS 2008 for C++, it's free on the microsoft site ) and try your code in that one. VS 6.0 has a LOT of bugs and even if the 2008 version still isn't working for you it's still an upgrade you MUST make!

One more thing, when you're testing the values, I hope you're in debug and looking directly at them! If you're printing them out on the console ... all bets are off :)

Could you please tell me what were the values of PD? Because I tried it on both VC++ 6 and 9..and both the versions i have the same problem. My value is 7.44061*10^7, the value remains the same for the whole for loop. Were you getting different values throughout? I am really confused in this regard.

Copied and pasted and the value I got was : 0.0004254 for PD

Could you please tell me what were the values of PD? Because I tried it on both VC++ 6 and 9..and both the versions i have the same problem. My value is 7.44061*10^7, the value remains the same for the whole for loop. Were you getting different values throughout? I am really confused in this regard.

Hi ,

I went through the code that I posted, I made a mistake in typing the lambda value. `lambda=532*10^-07` You are getting the value of PD different in every case because the k term is no longer 10^7 value. `k=1.18105*10^7` r values in the code are of the order 10^-05, the r values are negligible when we add it to 36.96 and then we when we multiply with a value of the order 10^07, there is a significant difference in values if we do not consider that truncation as well.

This is what I was asking about. How can I avoid the truncation that occurs when I do the step `r[n][m]+pinDrec` ?

Could somebody please help me with this problem?

Could you post a short, compilable code example where I could see the problem ?

Could you post a short, compilable code example where I could see the problem ?

My code is given below. PD value is where I have my problem.

``````Nx=1280;
Ny=960;
double pinhole_ccd_D_rec=6.3;
double lambda=5.32*10^-07
double pinDrec=pow(pinhole_ccd_D_rec,2);
double k=2*pi/lamda;//[1.18105e+07]
for (m=0;m<Nx;m++)
{ for(n=0;n<Ny;n++)
{ 	x[n][m]=(1+m-Nx/2); y[n][m]=(1+n-Ny/2);
r[n][m]= pow((x[n][m]*dx),2)+ pow((y[n][m]*dy),2);
PD=k*sqrt(r[n][m]+pinDrec);
}
}``````

thank you.

``e lambda=5.32*10^-07``

You do know that '^' is called XOR and is not "to the power of"
operator ?

10 ^ -07 != 0.0000001;

``e lambda=5.32*10^-07``

You do know that '^' is called XOR and is not "to the power of"
operator ?

10 ^ -07 != 0.0000001;

yes i do know..

``````Nx=1280;
Ny=960;
double pinhole_ccd_D_rec=6.3;
double lambda=5.32*pow(10,-7);
double pinDrec=pow(pinhole_ccd_D_rec,2);
double k=2*pi/lamda;//[1.18105e+07]
for (m=0;m<Nx;m++)
{ for(n=0;n<Ny;n++)
{ 	x[n][m]=(1+m-Nx/2); y[n][m]=(1+n-Ny/2);
r[n][m]= pow((x[n][m]*dx),2)+ pow((y[n][m]*dy),2);
PD=k*sqrt(r[n][m]+pinDrec);
}
}``````

Is it possible to post your complete code, if its not to big?

Yes, please post a short test code that compiles.

Be a part of the DaniWeb community

We're a friendly, industry-focused community of 1.21 million developers, IT pros, digital marketers, and technology enthusiasts learning and sharing knowledge.