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From: pete on 19 May 2010 02:57
amelia jane wrote:
>
> On May 17, 8:22 pm, pete <pfil...(a)mindspring.com> wrote:
> > amelia jane wrote:
> >
> > > Hello folks,
> >
> > > I have the following pseudo code:
> >
> > > #define MY_DEF 10
> > > void my_func(void)
> > > {
> >
> > > int i, loopnumber = 0; loopcount = 0, offset = 6;
> >
> > > loopnumber = (MY_DEF-offset)/2;
> > > for (i = 0; i< loopnumber; i++)
> > > {
> > > loopcount++;
> > > }
> >
> > > printf("loopcount=%d\n",loopcount);
> > > }
> >
> > > This code prints out loopcount=2.
> >
> > > I am a bit confused on why this is the case.
> > > Any thoughts on this are
> > > welcome.
> >
> > loopnumber is assigned a value of two.
> > The loop executes twice.
> > loopcount starts at zero and increments twice.

> Thanks for the responses so far. For the benefit of everyone, here is
> the original piece of code in C:
>
> #define MY_DEF 10
>
> void my_func(void)
> {
> int i;
> int loopcount;
> int offset = 6;
> loopcount = 0;
>
> for (i = 0; i< ((MY_DEF-offset)/2); i++)
> {
> loopcount++;
> }
>
> printf("loopcount=%d\n",loopcount);
> }
>
> main()
> {
> my_func();
> }
>
> I still don't get why loopcount equals 1;

It doesn't.
When I run that code, the output is:

loopcount=2

--
pete
From: Dann Corbit on 19 May 2010 03:21
In article <4BF38BF1.611E(a)mindspring.com>, pfiland(a)mindspring.com
says...
>
> It doesn't.
> When I run that code, the output is:
>
> loopcount=2

It could conceivably be anything. A varadic function (printf) is used
with no prototype in scope, and so undefined behavior is created.

If this program also does not produce 2 on the O.P.'s machine, then I am
at a loss for any possible explanation:

#include <stdio.h>

#define MY_DEF 10

void my_func(void)
{
int i;
int loopcount;
const int offset = 6;
loopcount = 0;

for (i = 0; i< ((MY_DEF-offset)/2); i++)
{
loopcount++;
}

printf("loopcount=%d\n",loopcount);
}

int main(void)
{
my_func();
return 0;
}
From: bart.c on 19 May 2010 05:56
amelia jane wrote:

> Thanks for the responses so far. For the benefit of everyone, here is
> the original piece of code in C:
>
> #define MY_DEF 10
>
> void my_func(void)
> {
> int i;
> int loopcount;
> int offset = 6;
> loopcount = 0;
>
> for (i = 0; i< ((MY_DEF-offset)/2); i++)
> {
> loopcount++;
> }
>
> printf("loopcount=%d\n",loopcount);
> }
>
> main()
> {
> my_func();
> }
>
> I still don't get why loopcount equals 1;

Put a printf call within the loop body (printing a message, and perhaps the
value of i, loopcount and the complete MY_DEF expression); then you can see
how many times the loop is executed. And maybe print the value of i after
the loop too; it should be 2.

And, is it really the original code? I think a superfluous semicolon at the
end of the 'for' line would give loopcount=1.

--
Bartc

From: Patricia Shanahan on 19 May 2010 06:52
amelia jane wrote:
> On May 17, 8:22 pm, pete <pfil...(a)mindspring.com> wrote:
>> amelia jane wrote:
>>
>>> Hello folks,
>>> I have the following pseudo code:
>>> #define MY_DEF 10
>>> void my_func(void)
>>> {
>>> int i, loopnumber = 0; loopcount = 0, offset = 6;
>>> loopnumber = (MY_DEF-offset)/2;
>>> for (i = 0; i< loopnumber; i++)
>>> {
>>> loopcount++;
>>> }
>>> printf("loopcount=%d\n",loopcount);
>>> }
>>> This code prints out loopcount=2.
>>> I am a bit confused on why this is the case. Any thoughts on this are
>>> welcome.
>> loopnumber is assigned a value of two.
>> The loop executes twice.
>> loopcount starts at zero and increments twice.
>>
>> --
>> pete
>
> Thanks for the responses so far. For the benefit of everyone, here is
> the original piece of code in C:
>
> #define MY_DEF 10
>
> void my_func(void)
> {
> int i;
> int loopcount;
> int offset = 6;
> loopcount = 0;
>
> for (i = 0; i< ((MY_DEF-offset)/2); i++)
> {
> loopcount++;
> }
>
> printf("loopcount=%d\n",loopcount);
> }
>
> main()
> {
> my_func();
> }
>
> I still don't get why loopcount equals 1;

I compiled the code with gcc and ran it, result "loopcount=2". Are you
sure this is exactly (copy and paste, no retyping) the code for which
you get loopcount=1? As Bartc suggested, the following program does get
"loopcount=1":

#define MY_DEF 10

void my_func(void)
{
int i;
int loopcount;
int offset = 6;
loopcount = 0;

for (i = 0; i< ((MY_DEF-offset)/2); i++);
{
loopcount++;
}

printf("loopcount=%d\n",loopcount);
}

main()
{
my_func();
}



Patricia

From: stan on 19 May 2010 13:01
Patricia Shanahan wrote:
> amelia jane wrote:
>> On May 17, 8:22 pm, pete <pfil...(a)mindspring.com> wrote:
>>> amelia jane wrote:
<snip>
>> Thanks for the responses so far. For the benefit of everyone, here is
>> the original piece of code in C:
>>
>> #define MY_DEF 10
>>
>> void my_func(void)
>> {
>> int i;
>> int loopcount;
>> int offset = 6;
>> loopcount = 0;
>>
>> for (i = 0; i< ((MY_DEF-offset)/2); i++)
>> {
>> loopcount++;
>> }
>>
>> printf("loopcount=%d\n",loopcount);
>> }
>>
>> main()
>> {
>> my_func();
>> }
>>
>> I still don't get why loopcount equals 1;
>
> I compiled the code with gcc and ran it, result "loopcount=2". Are you
> sure this is exactly (copy and paste, no retyping) the code for which
> you get loopcount=1? As Bartc suggested, the following program does get
> "loopcount=1":

As mentioned, after adding the stdio.h, I get the expected result.

I agree with the others as to possible causes, but I might add
that this may be a perfect time to learn to step through code in a
debugger. You might discover the problem with your code and you will
certainly learn a skill which can be very handy. It won't be time
wasted. Debuggers are not a silver bullet, they can't solve every
problem all the time. But when you fine yourself stumped you can never
have too many tools available.

Just a thought.
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