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Official Forum for Programming in Objective-C (the iPhone Programming Language) - Stephen Kochan
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Chapter 7
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fundamental problem / dealing with program 7.6
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Topic: fundamental problem / dealing with program 7.6 (Read 2969 times)
SiriusA
Newbie
Posts: 22
fundamental problem / dealing with program 7.6
«
on:
July 17, 2009, 12:05:47 PM »
I'm having a fundamental issue with program 7.6 and after wracking my brains I think I've pinned the issue to a problem with the statement (in 7.6) that reads "numerator /= u;" -- which keeps coming up as 0 when I run it.
I've replicated the issue in the following code program:
#import <Foundation/Foundation.h>
int main (int argc, const char * argv[]) {
NSAutoreleasePool * pool = [[NSAutoreleasePool alloc] init];
long double thing;
thing = .125;
NSLog (@"if we set it: %Lg", thing);
thing = 1/8;
NSLog (@"but if we divide: %Lg", thing);
[pool drain];
return 0;
}
here is what I'm getting:
2009-07-17 11:58:00.568 test2[6588:10b] if we set it: 0.125
2009-07-17 11:58:00.572 test2[6588:10b] but if we divide: 0
my expectation is that in both cases "thing" should be set to 0.125 -- but it's not when doing the equation. I'm convinced I'm overlooking something VERY simple in all of this but I can't figure out what.
Thanks for any help!
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skochan
Administrator
Hero Member
Posts: 3114
Re: fundamental problem / dealing with program 7.6
«
Reply #1 on:
July 17, 2009, 12:16:11 PM »
Please go back to Chapter 4 and reread the discussion on integer arithmetic (pp 58-60).
Cheers,
Steve Kochan
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SiriusA
Newbie
Posts: 22
Re: fundamental problem / dealing with program 7.6
«
Reply #2 on:
July 17, 2009, 12:28:52 PM »
thank you for your fast reply! I'm ashamed I forgot that section (as it had stood out at the time).
however, I did think about this when trying to diagnose the issue and had replaced my int declarations with double long as, on page 55, you show that a double long maybe "12.341".
does this mean that a double long may HOLD a decimal place value but that any arithmetic must be done under a float in order to manage decimal places while the math's being performed?
(I hope I'm not tiring your patience!)
BTW: LOVE The book.
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rgronlie
Global Moderator
Full Member
Posts: 212
Re: fundamental problem / dealing with program 7.6
«
Reply #3 on:
July 17, 2009, 01:25:05 PM »
When doing arithmetic the compiler will implicitly convert the numbers to the highest order of number in the equation.
It doesn't care about the type of variable the result will be stored in while performing the arithmetic. It will do an implicit conversion (pg. 62) to match the type of the variable
after
the arithmetic has been done.
(
int
/
int
) generates an
int
(
float
/
int
) generates a
float
(
int
/
float
) generates a
float
(
int
/
double
) generates a
double
(
float
/
double
) generates a
double
1 =
integer
1.0 =
double
1.0f =
float
1 / 8 = 0 //
only integers present so only integer math is performed
1.0 / 8 = 0.125 //
contains a floating point number (a
double
)so floating point math is performed
or you could explicitly typecast it
(double)1 / 8 = 0.125
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Sanity: Minds are like parachutes. Just because you've lost yours doesn't mean you can borrow mine.
SiriusA
Newbie
Posts: 22
Re: fundamental problem / dealing with program 7.6
«
Reply #4 on:
July 17, 2009, 04:31:56 PM »
ha!
I had written an extensive response to rgonlie -- loaded with thanks for trying to enlighten me -- further describing my confusion when I hypothesized and then tested a theory that seemed ridiculous. my proof was to change
thing = 1/8;
to
thing = 1.0 / 8.0;
and voila! it worked! ("thing = (float) 1 / (float) 8;" also worked).
this just seems so counter-intuitive to me. I would imagine that the compiler would take the numbers 1 and 8 (admittedly, ints as wrintten), do the math (resulting in a float) and then down-convert the result to fit the variable in which the result is to be stored. in other words, were the "receiving" variable an into, it would shoe-horn 0.125 into it as 0. I had assumed that the truncation comes at the end of this process but my experiment suggests it comes at the beginning. (see below)
I'd appreciate any further wise words as this does seem to contradict the book and rgonlie's response. for the moment, however, I seem to understand where to look if the results are not as expected.
#import <Foundation/Foundation.h>
int main (int argc, const char * argv[]) {
NSAutoreleasePool * pool = [[NSAutoreleasePool alloc] init];
int intVar;
float floatVar;
intVar = (float) 1 / (float) 8;
NSLog (@" intVar = %i", intVar);
floatVar = (float) 1 / (float) 8;
NSLog (@" floatVar = %f", floatVar);
floatVar = 1 / 8;
NSLog (@" floatVar = %f", floatVar);
[pool drain];
return 0;
}
2009-07-17 16:31:30.532 test2[6973:10b] intVar = 0
2009-07-17 16:31:30.534 test2[6973:10b] floatVar = 0.125000
2009-07-17 16:31:30.536 test2[6973:10b] floatVar = 0.000000
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SiriusA
Newbie
Posts: 22
Re: fundamental problem / dealing with program 7.6
«
Reply #5 on:
July 17, 2009, 04:36:12 PM »
ok, I've answered my own question (moments later) -- or perhaps you were trying to explain this all along, dunno.
p.63 reads "whenever to operands in an expression are integers (and this applies to short, unsigned, and long integers as well), the operation is carried out under the rules of integer arithmetic."
that's the short and the long of it. so, the compiler makes a (smart) choice. it sees integers and says "I'll save time/space/money/energy by doing this all as integer arithmetic."
makes sense to me now. sorry to be so daft (and to have forgotten what I read earlier in the week.
thanks!
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