DEVELOP > C > Some newbie casting issues
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DEVELOP > C |
| User: |
"SabaUd" |
| Date: |
01 Dec 2007 05:39:50 AM |
| Object: |
Some newbie casting issues |
Hello,
I wrote a short program that needs to calculate capacity of stack.
i have 2 ints and im trying to do so.
my code:
Stack* selectedStack = ...;
double fActual = (double)selectedStack->nActualSize;
double fAllocated = (double)selectedStack->nAllocatedSize;
double fCapacity = (fActual / fAllocated) * 100.0;
printf("%d ", (int)ceil(fCapacity));
Now - this one works, but it's disgusting code...
I'm sure there's a more elegant way to execute it...
any ideas?
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| User: "Peter Nilsson" |
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| Title: Re: Some newbie casting issues |
01 Dec 2007 10:27:31 PM |
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SabaUd <Saba...@gmail.com> wrote:
I wrote a short program that needs to calculate capacity
of stack.
i have 2 ints and im trying to do so.
my code:
Stack* selectedStack = ...;
double fActual = (double)selectedStack->nActualSize;
double fAllocated = (double)selectedStack->nAllocatedSize;
double fCapacity = (fActual / fAllocated) * 100.0;
printf("%d ", (int)ceil(fCapacity));
Now - this one works, but it's disgusting code...
Assuming nXXXXX are indeed integers...
printf("%3.0f%%\n", 100.0 * selectedStack->nActualSize
/ selectedStack->nAllocatedSize);
--
Peter
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| User: "Minimiscience" |
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| Title: Re: Some newbie casting issues |
01 Dec 2007 09:53:17 PM |
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SabaUd <SabaUdi@gmail.com> writes:
Hello,
I wrote a short program that needs to calculate capacity of stack.
i have 2 ints and im trying to do so.
my code:
Stack* selectedStack = ...;
double fActual = (double)selectedStack->nActualSize;
double fAllocated = (double)selectedStack->nAllocatedSize;
These two casts aren't actually needed, unless you're using a very strict
compiler or you're compiling this as C++ code.
double fCapacity = (fActual / fAllocated) * 100.0;
printf("%d ", (int)ceil(fCapacity));
This cast, however, is needed so that printf() is passed the right type of
value.
--
You can't write me, I :q!
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| User: "santosh" |
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| Title: Re: Some newbie casting issues |
01 Dec 2007 07:28:13 AM |
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SabaUd wrote:
Hello,
I wrote a short program that needs to calculate capacity of stack.
i have 2 ints and im trying to do so.
my code:
Stack* selectedStack = ...;
double fActual = (double)selectedStack->nActualSize;
double fAllocated = (double)selectedStack->nAllocatedSize;
double fCapacity = (fActual / fAllocated) * 100.0;
printf("%d ", (int)ceil(fCapacity));
Now - this one works, but it's disgusting code...
I'm sure there's a more elegant way to execute it...
any ideas?
Yes. Provide separate functions that take a Stack * argument and return
details (like capacity, size etc.) about it. And consider using size_t
as the object type for returning size and capacity related values,
since that is the natural type for such purposes in Standard C.
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| User: "SabaUd" |
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| Title: Re: Some newbie casting issues |
01 Dec 2007 05:52:54 AM |
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oh, another one I forgot:
I want to print "30%"...
how do I print '%'?
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| User: "Simon Gerber" |
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| Title: Re: Some newbie casting issues |
01 Dec 2007 06:02:35 AM |
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SabaUd @ Sat, 01 Dec 2007 03:52:54 -0800:
how do I print '%'?
by typing %%
--
Simon Gerber
simugerber (at) student (dot) ethz (dot) ch
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| User: "Richard Heathfield" |
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| Title: Re: Some newbie casting issues |
01 Dec 2007 06:23:18 AM |
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SabaUd said:
oh, another one I forgot:
I want to print "30%"...
how do I print '%'?
See page 13 of "The C Programming Language", 2nd edition, by Kernighan and
Ritchie.
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999
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