Floating Point

[geertjoos]

Hi,

A pice of code.

float A = 10;
float B = 7;
float result;

result = A / B;

when i run this program, i've got as result 1,4285714.... etcetc.
is it possible to get only two or one decimals after the floating point? And how?

Thanks in advance.
Geert

[Bob Smith]

"ge*******@pandora.Be" wrote:

Hi,

A pice of code.

float A = 10;
float B = 7;
float result;

result = A / B;

when i run this program, i've got as result 1,4285714.... etcetc.
is it possible to get only two or one decimals after the floating point? And how?

Thanks in advance.
Geert

round it
/B

[Geert Joos]

"Bob Smith" <bo*******@marketweighton.com> schreef in bericht
news:3F***************@marketweighton.com...

"ge*******@pandora.Be" wrote:

Hi,

A pice of code.

float A = 10;
float B = 7;
float result;

result = A / B;

when i run this program, i've got as result 1,4285714.... etcetc.
is it possible to get only two or one decimals after the floating point? And how?
Thanks in advance.
Geert
round it
/B

And how do you do that, can you give me an example? I'm just a bit new in
C++
Thanks.
Geert

[Adam Fineman]

Geert Joos wrote:

"Bob Smith" <bo*******@marketweighton.com> schreef in bericht

round it
/B

And how do you do that, can you give me an example? I'm just a bit new in
C++

Here's one way:

#include <iostream>
#include <cmath>

using namespace std;

int
main()
{
float A = 10;
float B = 7;

float result = A / B;

float prec = 2.0;
float shift_factor = pow(10, prec);

cout << round(result * shift_factor) / shift_factor << '\n';

return 0;
}

Note that the above method is not particularly robust, elagent, or
efficient. It's probably not even accurate for arbitrary values of
'result'. Use at your own risk.

- Adam

--
Reverse domain name to reply.

[Alexandros]

ge*******@pandora.Be escribió:

Hi,

A pice of code.

float A = 10;
float B = 7;
float result;

result = A / B;

when i run this program, i've got as result 1,4285714.... etcetc.
is it possible to get only two or one decimals after the floating point? And how?

Thanks in advance.
Geert

If what you want is to display the number with 2 decimals although it
actually has more you can do the following:
#include <iostream>
#include <iomanip>
using namespace std;

int main()
{
float a=3.239487;
float b=23.234;
float r=b/a;

cout << setprecision(3) << r << endl;

return 0;
}

[Jeff Schwab]

ge*******@pandora.Be wrote:

Hi,

A pice of code.

float A = 10;
float B = 7;
float result;

result = A / B;

when i run this program, i've got as result 1,4285714.... etcetc.
is it possible to get only two or one decimals after the floating point? And how?

Thanks in advance.
Geert

#include <iostream>

/* Truncate a number to (at most) two digits past the decimal point.
* NB: This implementation is naive.
*/
template< typename T >
T round2( T const& t )
{
return int( t * 100 ) / 100.0;
}

int main( )
{
float A = 10;
float B = 7;
float result;

result = A / B;

std::cout << result << '\n';
std::cout << round2( result ) << '\n';
}

[Walter]

"ge*******@pandora.Be" <ge*******@Pandora.Be> wrote in message
news:d7**************************@posting.google.c om...

float A = 10;
float B = 7;
float result;

result = A / B;

when i run this program, i've got as result 1,4285714.... etcetc.
is it possible to get only two or one decimals after the floating point?

And how?

float A = 10;
float B = 7;
float result;

result = ((long)((A * 100) / B)) / 100.0;

-Walter
www.digitalmars.com free C/C++/D compilers