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Junky · 11-22-2009, 08:08 AM

Code:

#include <stdio.h>
#include <stdlib.h>

int main(void) {
    double x; int i;
    scanf("%lg",&x);
    
    for (i = 1; i <= 4; i++) {
         x *= 10;
         printf("x = %lg\n",x);
         printf("int(x) = %d\n",int(x));
    }
    system("pause");
    return 0;    
}

This is just a part from some program, but the problem is here.
If someone could run it with x = 0.186, my question will be obvious

Thank you.

quzah · 11-22-2009, 08:21 AM

Why don't you just ask us your question instead of hoping someone comes along and compiles your code, runs it with the example input, and guesses as to what you think it's supposed to do?

Quzah.

**laserlight** · 11-22-2009, 08:22 AM

Quote:

Originally Posted by Junky

If someone could run it with x = 0.186, my question will be obvious

Well, your program cannot even compile, though that is easily fixed

I suggest that you explicitly state your question anyway.

Junky · 11-22-2009, 08:24 AM

Ok, sorry, i didn't mean to be rude.
I have x = 0.168, and I multiply it with 1000 (x *= 1000), and I get x = 186, but int(x) = 185.

But it compiles to me. I don't understand...

**laserlight** · 11-22-2009, 08:27 AM

Quote:

Originally Posted by Junky

I have x = 0.168, and I multiply it with 1000 (x *= 1000), and I get x = 186, but int(x) = 185.

That is because of floating point inaccuracy: although you entered 0.186, the actual value stored is a little less than that.

Junky · 11-22-2009, 08:29 AM

I just saw that if I put x as float instead of double (and scanf("%lg",&x) -> scanf("%f",&x)) it works fine.
But doesn't double have more precision than normal float?

Tnx for the explanation.

**laserlight** · 11-22-2009, 09:04 AM

Quote:

Originally Posted by Junky

I just saw that if I put x as float instead of double (and scanf("%lg",&x) -> scanf("%f",&x)) it works fine.

For this particular value; maybe it does not work on others.

Quote:

Originally Posted by Junky

But doesn't double have more precision than normal float?

It is about accuracy, not precision. As long as you cannot accurately represent a given value, it means that you only have an approximation, thus different approximations can give you different results. Perhaps what you are observing is a case where with more precision, a closer approximation was used, but this approximation is less than the actual value, whereas with less precision, the approximation is greater than the actual value.

Junky · 11-29-2009, 08:26 AM

I mostly understood what you said, tnx for explaining

Do you know if perhaps there is a way to do this correct? Is there a way to get a correct number with digits "abcd" from "0.abcd" (a way to bypass that error of not enough accuracy)?

Adak · 11-29-2009, 11:04 AM

yes, of course.

You use two integers to represent the two halves of a float or double number. Then you do your computations, on each int, as you want. Now every integer value can be accurately represented.

11-22-2009, 08:08 AM	#1
Junky Registered User Join Date: Nov 2009 Posts: 14	Help Code: #include <stdio.h> #include <stdlib.h> int main(void) { double x; int i; scanf("%lg",&x); for (i = 1; i <= 4; i++) { x *= 10; printf("x = %lg\n",x); printf("int(x) = %d\n",int(x)); } system("pause"); return 0; } This is just a part from some program, but the problem is here. If someone could run it with x = 0.186, my question will be obvious Thank you.
Junky is offline

11-22-2009, 08:21 AM	#2
quzah +++ OK NO CARRIER Join Date: Oct 2001 Posts: 10,640	Why don't you just ask us your question instead of hoping someone comes along and compiles your code, runs it with the example input, and guesses as to what you think it's supposed to do? Quzah. __________________ Hundreds of thousands of dipshits can't be wrong. Are you up for the suck?
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11-22-2009, 08:24 AM	#4
Junky Registered User Join Date: Nov 2009 Posts: 14	Ok, sorry, i didn't mean to be rude. I have x = 0.168, and I multiply it with 1000 (x = 1000), and I get x = 186, but int(x) = 185. But it compiles to me. I don't understand... Last edited by Junky; 11-22-2009 at 08:27 AM.*
Junky is offline

11-22-2009, 08:29 AM	#6
Junky Registered User Join Date: Nov 2009 Posts: 14	I just saw that if I put x as float instead of double (and scanf("%lg",&x) -> scanf("%f",&x)) it works fine. But doesn't double have more precision than normal float? Tnx for the explanation. Last edited by Junky; 11-22-2009 at 08:45 AM.
Junky is offline

11-29-2009, 08:26 AM	#8
Junky Registered User Join Date: Nov 2009 Posts: 14	I mostly understood what you said, tnx for explaining Do you know if perhaps there is a way to do this correct? Is there a way to get a correct number with digits "abcd" from "0.abcd" (a way to bypass that error of not enough accuracy)?
Junky is offline

11-29-2009, 11:04 AM	#9
Adak Registered User Join Date: Sep 2006 Posts: 3,151	yes, of course. You use two integers to represent the two halves of a float or double number. Then you do your computations, on each int, as you want. Now every integer value can be accurately represented.
Adak is offline

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