0 isn't 0 in C++ math

[justutiz]

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.

/*
 */
#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;
}

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
----------------------------------------------------------

[VictorN]

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.

... 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

[justutiz]

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.

[VictorN]

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

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:

if(abs(a - c) < delta)
{
   ...
}

[justutiz]

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

[VictorN]

Well, I meant this abs, _abs64

However, you can use fabs instead.

[Igor Vartanov]

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

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.

[justutiz]

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?

[Igor Vartanov]

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:

#include <stdio.h>

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

D:\Temp\8>8.exe

-0.000000

[justutiz]

Im still geting

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.

#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;
}

[Igor Vartanov]

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

[justutiz]

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

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

[justutiz]

so

if ( fabsf(c - a) < 0.000001 )

is the best and shortest way?

[VictorN]

so

if ( fabsf(c - a) < 0.000001 )

is the best and shortest way?

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.

[justutiz]

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?