Some newbie casting issues

[SabaUd]

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?

 

[SabaUd]

oh, another one I forgot:
I want to print "30%"...
how do I print '%'?

 

[Simon Gerber]

SabaUd @ Sat, 01 Dec 2007 03:52:54 -0800:

how do I print '%'?

by typing %%

 

[Richard Heathfield]

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.

 

[santosh]

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.

 

[Minimiscience]

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.

 

[Peter Nilsson]

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);