0 isn't 0 in C++ math

OK, i have a script, which calculate Y=1/(C-A). C is input, A is intetrval [Ap;Ag] with step Az;
And i have one problem, when c=2, A[-1.5;2.5] with step 0.1, when i want to say that 0 isn't possible in 1/0, the program colculate me: Y = -2.1e+06 . Programm is not English language, sorry. Problem in answer line 36.

Code:

---

/*
 */
#include <iostream>
#include <cstdlib>
#include <iomanip>
using namespace std;
int main ()
{
    float y, c, ap, az, ag;

        cout << "Iveskite kintamojo C reiksme.\n";
        cin >> c; //įvedame C reikšmę
        do {
                cout << "Iveskite intervalo pradzia ir gala.\nPradzia ir galas sutapti negali!\n";
            cin >> ap >> ag; //įvedame intervalo pradžią ir galą
        } while (ap == ag); //išsireikalaujame, kad intervalo pradžia ir galas nesutaptų 
          do {
                cout << "Iveskite intervalo kitimo zingsni, kuris nelygus 0.\n";
                cin >> az; //įvedame intervalo kitimo žingsnį
        } while (az == 0.0); //išsireikalaujame, kad intervalo kitimo ˛ingsnis būtų nelygus 0
        if (ag > ap && az > 0.0) { //jei intervalas didėjantis IR žingsnis DIDESNIS už 0, skaičiuojame
                cout << setfill('-') << setw(58) << "-" << endl;
                cout << setfill(' ') << setw(8)<<"Sprendinio nr. " <<"|  "<<"Atsakymas\n";
                cout << setfill('-') << setw(58) << "-" << setfill(' ') <<endl;
                for (float z = 1.0, a = ap; a <= ag; a += az) {
                        if (a == c) { //jei vardiklis lygus 0
                                cout << setfill(' ') << setw(14) << z << ".  ";
                cout << "Kai A = " << a <<", o " << c << " - " << a <<" = 0, apskaiciuoti\nnegalima, nes dalyba is 0 negalima.\n";
                        }
                        else {
                                y = 1.0 / (c - a); //skaičiuojame
                                cout << setfill(' ') << setw(14) << z << ".  " << "Y = "  << setprecision(3) << y << endl;       
                        }
                        z++; //prisideame vienetuką prie sprendinio nr.
                }       
       
        }
        if (ag < ap && az < 0.0) { //jei intervalas mažėjantis IR žingsnis MAŽESNIS uz 0
                cout << setfill('-') << setw(58) << "-" << endl;
                cout << setfill(' ') << setw(8)<<"Sprendinio nr. " <<"|  "<<"Atsakymas\n";
                cout << setfill('-') << setw(58) << "-" << setfill(' ') <<endl;
                for (float z = 1.0, a = ap; a >= ag; a += az) {
                        if (a == c) { //jei vardiklis lygus 0
                                cout << setfill(' ') << setw(14) << z << ".  ";
                cout << "Kai A = " << a <<", o " << c << " - " << a <<" = 0, apskaiciuoti\nnegalima, nes dalyba is 0 negalima.\n";
                        }
                        else {
                                y = 1.0 / (c - a); //skaičiuojame
                                cout << setfill(' ') << setw(14) << z << ".  " << "Y = "  << setprecision(3) << y << endl;       
                        }
                        z++; //prisideame vienetuką prie sprendinio nr.
                }       
               
        }
        if (ag > ap && az < 0.0) { //jei intervalas didėjantis IR žingsnis MAŽESNIS uz 0
                cout << setfill('-') << setw(58) << "-" << endl;
                cout << "Apskaiciuoti negalime,\nnes intervalas didejantis,\no zingsnis mazejantis.\n";
        }                       
        if (ag < ap && az > 0.0) { //jei intervalas mažejantis IR žingsnis DIDESNIS už 0
                cout << setfill('-') << setw(58) << "-" << endl;
                cout << "Apskaiciuoti negalime,\nnes intervalas mazejantis,\no zingsnis didejantis.\n";
        }                       
        cout << setfill('-') << setw(58) << "-" << setfill(' ') <<endl;
        system ("pause");
        return 0;
}

---

Code:

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Iveskite kintamojo C reiksme.
2
Iveskite intervalo pradzia ir gala.
Pradzia ir galas sutapti negali!
-1.5
2.5
Iveskite intervalo kitimo zingsni, kuris nelygus 0.
0.1
----------------------------------------------------------
Sprendinio nr. |  Atsakymas
----------------------------------------------------------
            1.  Y = 0.286
            2.  Y = 0.294
            3.  Y = 0.303
            4.  Y = 0.313
            5.  Y = 0.323
            6.  Y = 0.333
            7.  Y = 0.345
            8.  Y = 0.357
            9.  Y = 0.37
            10.  Y = 0.385
            11.  Y = 0.4
            12.  Y = 0.417
            13.  Y = 0.435
            14.  Y = 0.455
            15.  Y = 0.476
            16.  Y = 0.5
            17.  Y = 0.526
            18.  Y = 0.556
            19.  Y = 0.588
            20.  Y = 0.625
            21.  Y = 0.667
            22.  Y = 0.714
            23.  Y = 0.769
            24.  Y = 0.833
            25.  Y = 0.909
            26.  Y = 1
            27.  Y = 1.11
            28.  Y = 1.25
            29.  Y = 1.43
            30.  Y = 1.67
            31.  Y = 2
            32.  Y = 2.5
            33.  Y = 3.33
            34.  Y = 5
            35.  Y = 10
            36.  Y = -2.1e+06
            37.  Y = -10
            38.  Y = -5
            39.  Y = -3.33
            40.  Y = -2.5
            41.  Y = -2
----------------------------------------------------------

---

Quote:

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Originally Posted by justutiz

OK, i have a script, which calculate ...

---

Scripts are not discussed in this Forum.
There is a Visual C++ Programming forum, and Visual C++ is code, not script.

Quote:

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Originally Posted by justutiz

... which calculate Y=1/(C-A). C is input, A is intetrval [Ap;Ag] with step Az;

---

Did you, perhaps, mean "A is a value within intetrval [Ap;Ag]"?

Anyway, have a look at:
Floating point
What Every Computer Scientist Should Know About Floating-Point Arithmetic
and a lot of other articles/discussions here

Quote:

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Originally Posted by VictorN

Scripts are not discussed in this Forum.
There is a Visual C++ Programming forum, and Visual C++ is code, not script.

Did you, perhaps, mean "A is a value within intetrval [Ap;Ag]"?

Anyway, have a look at:
Floating point
What Every Computer Scientist Should Know About Floating-Point Arithmetic
and a lot of other articles/discussions here

---

Im reading, and Im very new in C++, so I dont find how I can do the folowing that a == c, then a and c is floats.

Quote:

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Originally Posted by justutiz

... so I dont find how I can do the folowing that a == c, then a and c is floats.

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In general you cannot just compare floats as "a == c". You have to implement some relative small (with respect to the "a" and "c") value ("delta") and compare like:

Code:

---

if(abs(a - c) < delta)
{
  ...
}

---

but abs want me to use int, my a and c is floats

Well, I meant this abs, _abs64

However, you can use fabs instead.

Quote:

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when i want to say that 0 isn't possible in 1/0, the program colculate me: Y = -2.1e+06

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Result of 1/0 is infinity, while programming languages operate with only finite figures. So you have the best result FPU was able to provide. :)

Im still geting

Code:

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1      -1      0
1.11      -0.9      0
1.25      -0.8      0
1.43      -0.7      0
1.67      -0.6      0
2      -0.5      0
2.5      -0.4      0
3.33      -0.3      0
5      -0.2      0
10      -0.1      0
-1.34e+07      7.45e-08      0
-10      0.1      0
-5      0.2      0
-3.33      0.3      0
-2.5      0.4      0
-2      0.5      0
-1.67      0.6      0
-1.43      0.7      0
-1.25      0.8      0
-1.11      0.9      0

---

trying everything, and still dont get how to fix it.

Code:

---

#include <iostream>
#include <cstdlib>
#include <iomanip>
#include <math.h>

using namespace std;
int main ()
{
    float y, c, ap, ag;
    c = 0;
    ap = -1;
    ag = 1;
    float az = 0.1;
    if (ag > ap && az > 0.0) {               
                for (float a = ap; a <= ag; a += az) {
                        if ( fabsf(c - a) < 0.0 && fabsf(c - a) > 0.0 ) {
                cout << "1/0 not posible";
                        }
                        else {
                                y = 1.0 / (c - a);
                cout  << setprecision(4) << y <<"      " << a << "      " << c << endl;       
                        }
                }       
       
        }
    else cout << "STOP";
        system ("pause");
        return 0;
}

---

It's impossible to succeed with the condition fabsf(c - a) < 0.0. Victor meant something like fabsf(c - a) < 0.00001.

Quote:

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Originally Posted by Igor Vartanov

Result of 1/0 is infinity, while programming languages operate with only finite figures. So you have the best result FPU was able to provide. :)

---

Yes, but why in my code -0.1 + 0.1 is not 0?

Quote:

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Originally Posted by Igor Vartanov

It's impossible to succeed with the condition fabsf(c - a) < 0.0. Victor meant something like fabsf(c - a) < 0.00001.

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Yes, it works, but maybe it is shortes way to compare c-a is 0?

Quote:

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Originally Posted by justutiz

Yes, but why in my code -0.1 + 0.1 is not 0?

---

I don't know what's wrong with your code, but mine works fine:

Code:

---

#include <stdio.h>

int main()
{
    float a = -0.1;
    printf("%f", a +0.1);
    return 0;
}

---

Quote:

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D:\Temp\8>8.exe

-0.000000

---

Quote:

---

Originally Posted by Igor Vartanov

I don't know what's wrong with your code, but mine works fine:

Code:

---

#include <stdio.h>

int main()
{
    float a = -0.1;
    printf("%f", a +0.1);
    return 0;
}

---

---

when add one time its true, but when you do it with cycle it looks something like this

a = -0.1 + 0.01000000001 and so on
and then you compare 0 with a it is looks something like this 0 != 0.0000000000001

Quote:

---

Originally Posted by justutiz

Yes, it works, but maybe it is shortes way to compare c-a is 0?

---

Well, the big mistake you make is that you think math and programming are the same. Floats in programming have finite precision, so you must take certain precautions and tricks to succeed with your calculations. Like comparing with 0.00001, for example.

Quote:

---

Originally Posted by justutiz

when add one time its true, but when you do it with cycle it looks something like this

a = -0.1 + 0.01000000001 and so on
and then you compare 0 with a it is looks something like this 0 != 0.0000000000001

---

Yep, see above: float calculations have finite precision. And the result might fluctuate a little. :)

Quote:

---

Originally Posted by justutiz

when add one time its true, but when you do it with cycle it looks something like this

a = -0.1 + 0.01000000001 and so on
and then you compare 0 with a it is looks something like this 0 != 0.0000000000001

---

Haven't you read yet what I referred to in the post#2?

so

Code:

---

if ( fabsf(c - a) < 0.000001 )

---

is the best and shortest way?

Quote:

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Originally Posted by justutiz

so

Code:

---

if ( fabsf(c - a) < 0.000001 )

---

is the best and shortest way?

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I'd say it is the only way. The question is only about this 'delta' value. In some case your choice with 0.000001 can work, in other - not. It all depends on the problem / values you are working on / with.

Quote:

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Originally Posted by VictorN

I'd say it is the only way. The question is only about this 'delta' value. In some case your choice with 0.000001 can work, in other - not. It all depends on the problem / values you are working on / with.

---

so it is posible to make working with all values, if az i take 0.000000001 then it is not working, so i need to add more 0.00000000000000001 in compare, it is other way?

Again: did you read about FP arithmetic?

Quote:

---

Originally Posted by justutiz

so it is posible to make working with all values, if az i take 0.000000001 then it is not working, so i need to add more 0.00000000000000001 in compare, it is other way?

---

Before you ask more questions, please read this:

http://www.parashift.com/c++-faq-lit...html#faq-29.16
http://www.parashift.com/c++-faq-lit...html#faq-29.17
http://www.parashift.com/c++-faq-lit...html#faq-29.18

Computers work in binary, not decimal. This means that the decimal numbers you're trying to use must be converted to binary. Do you know how to convert a decimal floating point value to a binary floating point value? If you do, then you will see why decimal fractions cannot have exact binary representations unless they are a sum of inverse powers of 2.

If you want exact numbers, then don't use float or double. Use a class that represents arbitrary precision numbers, and there are many of them out there.

Regards,

Paul McKenzie

Quote:

---

Originally Posted by VictorN

Again: did you read about FP arithmetic?

---

Yes, but dont understand everything. I just need to do a program in few days. This is my first program and I am still learning everything.

Quote:

---

Originally Posted by justutiz

Yes, but dont understand everything. I just need to do a program in few days. This is my first program and I am still learning everything.

---

Well, this code will not work:

Code:

---

for (float a = ap; a <= ag; a += az)

---

Never use floats or doubles as loop counters. Never.

The reason is that you cannot guarantee how many times that loop will execute. It could execute 1 more time than expected, and the reason was explained by everyone so far -- floating point computations are not exact.

Always use integers as loop counters. This guarantees that the loop will execute the exact number of times you expect. For example, if you know you should loop from 0.01 to 0.1 then you write this:

Code:

---

for (int i = 1; i <= 100; ++i)
{
    double myValue = i / 100.0;
    // then you use myValue in the loop, not i
    //...
}

---

Now the loop is guaranteed to loop 100 times. The bottom line is however you do it, the goal is to normalize all "counted" loops to use integers as loop counters.

Regards,

Paul McKenzie