From:

Robert Wessel <robertwessel2@yahoo.com>

Newsgroups:

comp.lang.c++

Date:

Thu, 09 Jan 2014 23:39:05 -0600

Message-ID:

<h8vuc9p275oihr1sio3h5shmoc5brj5lk3@4ax.com>

wrote:

On 9/1/14 5:46 pm, Raymond Li wrote:

Thanks for your replies. I hope to stick to double too. But the users

have implemented the logic in legacy system and I need to convince them

if I do something different from them. They claimed that the interim

calculations (z[i]*weighted/average) are used too and they would feel

uncomfortable if I make any adjustment. The worst problem I faced is

that they claimed that the legacy system (which is not really legacy, it

is running Oracle pl/sql) does not have the rounding error.

So I investigated and found it weird that the rounding problem could be

avoided by using float. I am uncomfortable to this workaround (using

float), as I am afraid there would be cases that the rounding issue

recur in other scenarios. So I really want someone could explain why the

float datatype would round correctly in the above case, while using

double rounded 'incorrectly'.

I have encountered a problem related to floating point rounding. I

googled a lot and there are many clear and helpful information. e.g.

http://www.learncpp.com/cpp-tutorial/25-floating-point-numbers/

http://support.microsoft.com/kb/125056/en-hk

Although the urls have explained the cause, I need to find a practical

way to solve a rounding problem. My program has calculated a weighted

accumulation as 3.5. When the figure is rounded to nearest number, it

became 3 (but I want it to round up to 4). I understood it would be due

to approximation value of 3.5 as 3.49999...

I found a simple fix by using float instead of double. I list the

program below and wish someone could explain why using double would

incur the rounding problem while float would not. In the code below,

fun1() use float and the calculation is 'correct'. In fun2(), it uses

double and the figure 3.5 is rounded as 3.

Raymond

//######################

#include <cmath>

#include <iostream>

//using namespace std;

using std::cout;

using std::endl;

int fun1();

int fun2();

int main(int argc, char ** argv)

{

fun1();

fun2();

return 0;

}

int fun1()

{

float weighted=10.0;

float average=100.0;

float z[]=

{

4.0,

4.0,

4.0,

4.0,

4.0,

3.0,

3.0,

3.0,

2.0,

4.0

};

float total=0.0;

int i=0;

for (i=0;i<10;i++)

{

float item=z[i]*weighted/average;

total=total+item;

cout << i << " accumulate is " << total << endl;

// NSLog(@"z[%d] is %f, total is %f", i, z[i], total);

}

float answer=round(total);

// NSLog(@"rounded is %f", answer);

cout << "rounded is " << answer << endl;

return 0;

}

int fun2()

{

double weighted=10.0;

double average=100.0;

double z[]=

{

4.0,

4.0,

4.0,

4.0,

4.0,

3.0,

3.0,

3.0,

2.0,

4.0

};

double total=0.0;

int i=0;

for (i=0;i<10;i++)

{

double item=z[i]*weighted/average;

total=total+item;

cout << i << " accumulate is " << total << endl;

// NSLog(@"z[%d] is %f, total is %f", i, z[i], total);

}

double answer=round(total);

// NSLog(@"rounded is %f", answer);

cout << "rounded is " << answer << endl;

return 0;

}

0 accumulate is 0.4

1 accumulate is 0.8

2 accumulate is 1.2

3 accumulate is 1.6

4 accumulate is 2

5 accumulate is 2.3

6 accumulate is 2.6

7 accumulate is 2.9

8 accumulate is 3.1

9 accumulate is 3.5

rounded is 4

***(above is the version using float, 3.5 is rounded as 4) ***

0 accumulate is 0.4

1 accumulate is 0.8

2 accumulate is 1.2

3 accumulate is 1.6

4 accumulate is 2

5 accumulate is 2.3

6 accumulate is 2.6

7 accumulate is 2.9

8 accumulate is 3.1

9 accumulate is 3.5

rounded is 3

***(this version use double, 3.5 is rounded as 3) ***

--- news://freenews.netfront.net/ - complaints: news@netfront.net ---

googled a lot and there are many clear and helpful information. e.g.

http://www.learncpp.com/cpp-tutorial/25-floating-point-numbers/

http://support.microsoft.com/kb/125056/en-hk

Although the urls have explained the cause, I need to find a practical

way to solve a rounding problem. My program has calculated a weighted

accumulation as 3.5. When the figure is rounded to nearest number, it

became 3 (but I want it to round up to 4). I understood it would be due

to approximation value of 3.5 as 3.49999...

I found a simple fix by using float instead of double. I list the

program below and wish someone could explain why using double would

incur the rounding problem while float would not. In the code below,

fun1() use float and the calculation is 'correct'. In fun2(), it uses

double and the figure 3.5 is rounded as 3.

Raymond

//######################

#include <cmath>

#include <iostream>

//using namespace std;

using std::cout;

using std::endl;

int fun1();

int fun2();

int main(int argc, char ** argv)

{

fun1();

fun2();

return 0;

}

int fun1()

{

float weighted=10.0;

float average=100.0;

float z[]=

{

4.0,

4.0,

4.0,

4.0,

4.0,

3.0,

3.0,

3.0,

2.0,

4.0

};

float total=0.0;

int i=0;

for (i=0;i<10;i++)

{

float item=z[i]*weighted/average;

total=total+item;

cout << i << " accumulate is " << total << endl;

// NSLog(@"z[%d] is %f, total is %f", i, z[i], total);

}

float answer=round(total);

// NSLog(@"rounded is %f", answer);

cout << "rounded is " << answer << endl;

return 0;

}

int fun2()

{

double weighted=10.0;

double average=100.0;

double z[]=

{

4.0,

4.0,

4.0,

4.0,

4.0,

3.0,

3.0,

3.0,

2.0,

4.0

};

double total=0.0;

int i=0;

for (i=0;i<10;i++)

{

double item=z[i]*weighted/average;

total=total+item;

cout << i << " accumulate is " << total << endl;

// NSLog(@"z[%d] is %f, total is %f", i, z[i], total);

}

double answer=round(total);

// NSLog(@"rounded is %f", answer);

cout << "rounded is " << answer << endl;

return 0;

}

0 accumulate is 0.4

1 accumulate is 0.8

2 accumulate is 1.2

3 accumulate is 1.6

4 accumulate is 2

5 accumulate is 2.3

6 accumulate is 2.6

7 accumulate is 2.9

8 accumulate is 3.1

9 accumulate is 3.5

rounded is 4

***(above is the version using float, 3.5 is rounded as 4) ***

0 accumulate is 0.4

1 accumulate is 0.8

2 accumulate is 1.2

3 accumulate is 1.6

4 accumulate is 2

5 accumulate is 2.3

6 accumulate is 2.6

7 accumulate is 2.9

8 accumulate is 3.1

9 accumulate is 3.5

rounded is 3

***(this version use double, 3.5 is rounded as 3) ***

--- news://freenews.netfront.net/ - complaints: news@netfront.net ---

Thanks for your replies. I hope to stick to double too. But the users

have implemented the logic in legacy system and I need to convince them

if I do something different from them. They claimed that the interim

calculations (z[i]*weighted/average) are used too and they would feel

uncomfortable if I make any adjustment. The worst problem I faced is

that they claimed that the legacy system (which is not really legacy, it

is running Oracle pl/sql) does not have the rounding error.

So I investigated and found it weird that the rounding problem could be

avoided by using float. I am uncomfortable to this workaround (using

float), as I am afraid there would be cases that the rounding issue

recur in other scenarios. So I really want someone could explain why the

float datatype would round correctly in the above case, while using

double rounded 'incorrectly'.

You shouldn't depend on that, it's just a coincidence of how the

rounding error happened to accumulate.

I've modified you program to display a bit more precision (attached

below). With the better display of precision, you can see the

roundoff errors accumulating differently:

(float) 0 item is 0.40000000596046448 accumulate is

0.40000000596046448

(float) 1 item is 0.40000000596046448 accumulate is

0.80000001192092896

(float) 2 item is 0.40000000596046448 accumulate is 1.2000000476837158

(float) 3 item is 0.40000000596046448 accumulate is 1.6000000238418579

(float) 4 item is 0.40000000596046448 accumulate is 2

(float) 5 item is 0.30000001192092896 accumulate is 2.2999999523162842

(float) 6 item is 0.30000001192092896 accumulate is 2.5999999046325684

(float) 7 item is 0.30000001192092896 accumulate is 2.8999998569488525

(float) 8 item is 0.20000000298023224 accumulate is 3.0999999046325684

(float) 9 item is 0.40000000596046448 accumulate is 3.5

rounded is 4

(double) 0 item is 0.40000000000000002 accumulate is

0.40000000000000002

(double) 1 item is 0.40000000000000002 accumulate is

0.80000000000000004

(double) 2 item is 0.40000000000000002 accumulate is

1.2000000000000002

(double) 3 item is 0.40000000000000002 accumulate is

1.6000000000000001

(double) 4 item is 0.40000000000000002 accumulate is 2

(double) 5 item is 0.29999999999999999 accumulate is

2.2999999999999998

(double) 6 item is 0.29999999999999999 accumulate is

2.5999999999999996

(double) 7 item is 0.29999999999999999 accumulate is

2.8999999999999995

(double) 8 item is 0.20000000000000001 accumulate is

3.0999999999999996

(double) 9 item is 0.40000000000000002 accumulate is

3.4999999999999996

rounded is 3

But as I said, you can depend on that. For example, changing the

series to:

{

9.0,

9.0,

9.0,

8.0,

0.0,

0.0,

0.0,

0.0,

0.0,

0.0

};

Will cause the float version to round to 3 as well:

(float) 0 item is 0.89999997615814209 accumulate is

0.89999997615814209

(float) 1 item is 0.89999997615814209 accumulate is 1.7999999523162842

(float) 2 item is 0.89999997615814209 accumulate is 2.6999998092651367

(float) 3 item is 0.80000001192092896 accumulate is 3.4999997615814209

(float) 4 item is 0 accumulate is 3.4999997615814209

(float) 5 item is 0 accumulate is 3.4999997615814209

(float) 6 item is 0 accumulate is 3.4999997615814209

(float) 7 item is 0 accumulate is 3.4999997615814209

(float) 8 item is 0 accumulate is 3.4999997615814209

(float) 9 item is 0 accumulate is 3.4999997615814209

rounded is 3

IOW, this will vary with the exact set of inputs. So don't do that.

Even worse, you can get this to wander around based on whether or not

you tell the compiler to produce strict IEEE compliant math, and the

requested optimization level (on x86 machines you often see

intermediate results with a higher precision than you'd expect if the

code is using the x87 FPU).

Changing the values (as has been suggested) so that all of them have

exact representations can reduce the roundoff error, but cannot

eliminate it (you'll still get roundoff error on that final division,

even if you get none on the individual terms). OTOH, you probably

will get away with this so long as the only case you care about is

"xxx5.0 / 10.0", since that will have an exact result (.5 being

exactly representable in a binary FP number). This is obviously

fragile, and will go to pot the first time someone tosses in a number

with more than a single decimal place.

If this is important, I'd generally advise avoiding floating point

entirely, and use a package that allows you to do this all with scaled

integers or rationals. I'm not sure what the PL/SQL code is doing,

but it may be using a scaled type, or if it's using a floating type,

they might just be hitting one set of rounding errors that happens to

generate the expected result. And that may just be the worst scenario

- trying to duplicate the existing behavior when the existing behavior

is not what anyone is actually expecting.

/* ----- */

#include <cmath>

#include <iostream>

#include <iomanip>

using std::cout;

using std::endl;

int fun1();

int fun2();

inline int round(double x) { return (floor(x + 0.5)); }

int main(int argc, char ** argv)

{

fun1();

fun2();

return 0;

}

int fun1()

{

float weighted=10.0;

float average=100.0;

float z[]=

{

4.0,

4.0,

4.0,

4.0,

4.0,

3.0,

3.0,

3.0,

2.0,

4.0

};

float total=0.0;

int i=0;

for (i=0;i<10;i++)

{

float item=z[i]*weighted/average;

total=total+item;

cout << "(float) " << i << " item is " <<

std::setprecision(20) << item

<< " accumulate is " << std::setprecision(20) << total

<< endl;

}

float answer=round(total);

cout << "rounded is " << answer << endl;

return 0;

}

int fun2()

{

double weighted=10.0;

double average=100.0;

double z[]=

{

4.0,

4.0,

4.0,

4.0,

4.0,

3.0,

3.0,

3.0,

2.0,

4.0

};

double total=0.0;

int i=0;

for (i=0;i<10;i++)

{

double item=z[i]*weighted/average;

total=total+item;

cout << "(double) " << i << " item is " <<

std::setprecision(20) << item

<< " accumulate is " << std::setprecision(20) << total

<< endl;

}

double answer=round(total);

cout << "rounded is " << answer << endl;

return 0;

}

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