Question on the FAQ

[Topi Linkala]

Question 3.15:

Why does the code

double degC, degF;
degC = 5 / 9 * (degF - 32);

keep giving me 0?


Would the following rearrangement solve the problem?

degC = 5 * (degF - 32) / 9;


Topi Linkala
--
"The whole problem with the world is that fools and fanatics are
always so certain of themselves, but wiser people so full of doubts."
- Bertrand Russell
"How come he didn't put 'I think' at the end of it?" - Anonymous

[Robert Gamble]

Re: Question on the FAQ

On Apr 13, 3:33 pm, Topi Linkala <n...@iki.fiwro te:

Question 3.15:
>
Why does the code
>
double degC, degF;
degC = 5 / 9 * (degF - 32);
>
keep giving me 0?
>
Would the following rearrangement solve the problem?
>
degC = 5 * (degF - 32) / 9;
>

Did you read the answer to the question (<http://c-faq.com/expr/
truncation1.htm l>)?

--
Robert Gamble

[Topi Linkala]

Re: Question on the FAQ

Robert Gamble wrote:

On Apr 13, 3:33 pm, Topi Linkala <n...@iki.fiwro te:
>

>>Question 3.15:
>>
>>Why does the code
>>
>>double degC, degF;
>>degC = 5 / 9 * (degF - 32);
>>
>>keep giving me 0?
>>
>>Would the following rearrangement solve the problem?
>>
>>degC = 5 * (degF - 32) / 9;
>>

>
>
Did you read the answer to the question (<http://c-faq.com/expr/
truncation1.htm l>)?

Yes I read and it says:

"If both operands of a binary operator are integers, C performs an
integer operation,..."

But in my rewriting there are no binary operation with integer values.
That's why I'm asking if it would work.

Topi
--
"The whole problem with the world is that fools and fanatics are
always so certain of themselves, but wiser people so full of doubts."
- Bertrand Russell
"How come he didn't put 'I think' at the end of it?" - Anonymous

[Robert Gamble]

Re: Question on the FAQ

On Apr 13, 3:48 pm, Topi Linkala <n...@iki.fiwro te:

Robert Gamble wrote:

On Apr 13, 3:33 pm, Topi Linkala <n...@iki.fiwro te:

>

>Question 3.15:

>

>Why does the code

>

>double degC, degF;
>degC = 5 / 9 * (degF - 32);

>

>keep giving me 0?

>

>Would the following rearrangement solve the problem?

>

>degC = 5 * (degF - 32) / 9;

>

Did you read the answer to the question (<http://c-faq.com/expr/
truncation1.htm l>)?

>
Yes I read and it says:
>
"If both operands of a binary operator are integers, C performs an
integer operation,..."
>
But in my rewriting there are no binary operation with integer values.
That's why I'm asking if it would work.
>

Sorry, I missed the fact that degF was a double. Yes, your example
will work properly since the division will be happening on floating
point values.

--
Robert Gamble

[Joe Wright]

Re: Question on the FAQ

Topi Linkala wrote:

Robert Gamble wrote:
>

>On Apr 13, 3:33 pm, Topi Linkala <n...@iki.fiwro te:
>>

>>Question 3.15:
>>>
>>Why does the code
>>>
>>double degC, degF;
>>degC = 5 / 9 * (degF - 32);
>>>
>>keep giving me 0?
>>>
>>Would the following rearrangement solve the problem?
>>>
>>degC = 5 * (degF - 32) / 9;
>>>

>>
>>
>Did you read the answer to the question (<http://c-faq.com/expr/
>truncation1.ht ml>)?

>
Yes I read and it says:
>
"If both operands of a binary operator are integers, C performs an
integer operation,..."
>
But in my rewriting there are no binary operation with integer values.
That's why I'm asking if it would work.
>
Topi

The statement..

degC = 5 / 9 * (degF - 32);

...can't work because 5 / 9 is (int) zero. Also degF is not initialized.

degC = (degF - 32) * 5 / 9;

should work better for you.

--
Joe Wright
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---

[Topi Linkala]

Re: Question on the FAQ

Joe Wright wrote:

Topi Linkala wrote:
>

>Robert Gamble wrote:
>>

>>On Apr 13, 3:33 pm, Topi Linkala <n...@iki.fiwro te:
>>>
>>>Question 3.15:
>>>>
>>>Why does the code
>>>>
>>>double degC, degF;
>>>degC = 5 / 9 * (degF - 32);
>>>>
>>>keep giving me 0?
>>>>
>>>Would the following rearrangement solve the problem?
>>>>
>>>degC = 5 * (degF - 32) / 9;
>>>>
>>>
>>>
>>Did you read the answer to the question (<http://c-faq.com/expr/
>>truncation1.h tml>)?

>>
>>
>Yes I read and it says:
>>
>"If both operands of a binary operator are integers, C performs an
>integer operation,..."
>>
>But in my rewriting there are no binary operation with integer values.
>That's why I'm asking if it would work.
>>
>Topi

>
>
The statement..
>
degC = 5 / 9 * (degF - 32);
>
..can't work because 5 / 9 is (int) zero. Also degF is not initialized.
>
degC = (degF - 32) * 5 / 9;
>
should work better for you.
>

I don't think so because there's no reason to believe that the
multiplication is done before the division, but in my example that
doesn't matter as one of the operands is double in any case.

Topi
--
"The whole problem with the world is that fools and fanatics are
always so certain of themselves, but wiser people so full of doubts."
- Bertrand Russell
"How come he didn't put 'I think' at the end of it?" - Anonymous

[Robert Gamble]

Re: Question on the FAQ

On Apr 13, 4:53 pm, Topi Linkala <n...@iki.fiwro te:

Joe Wright wrote:

Topi Linkala wrote:

>

Robert Gamble wrote:

>

>On Apr 13, 3:33 pm, Topi Linkala <n...@iki.fiwro te:

>

>>Question 3.15:

>

>>Why does the code

>

>>double degC, degF;
>>degC = 5 / 9 * (degF - 32);

>

>>keep giving me 0?

>

>>Would the following rearrangement solve the problem?

>

>>degC = 5 * (degF - 32) / 9;

>

>Did you read the answer to the question (<http://c-faq.com/expr/
>truncation1.ht ml>)?

>

Yes I read and it says:

>

"If both operands of a binary operator are integers, C performs an
integer operation,..."

>

But in my rewriting there are no binary operation with integer values.
That's why I'm asking if it would work.

>

Topi

>

The statement..

>

degC = 5 / 9 * (degF - 32);

>

..can't work because 5 / 9 is (int) zero. Also degF is not initialized.

>

degC = (degF - 32) * 5 / 9;

>

should work better for you.

>
I don't think so because there's no reason to believe that the
multiplication is done before the division, but in my example that
doesn't matter as one of the operands is double in any case.

Actually, the multiplication is guaranteed to be performed before the
division because they both have the same precedence and are left-to-
right associative.

--
Robert Gamble

[Flash Gordon]

Re: Question on the FAQ

Topi Linkala wrote, On 13/04/08 20:48:

Robert Gamble wrote:
>

>On Apr 13, 3:33 pm, Topi Linkala <n...@iki.fiwro te:
>>

>>Question 3.15:
>>>
>>Why does the code
>>>
>>double degC, degF;
>>degC = 5 / 9 * (degF - 32);
>>>
>>keep giving me 0?
>>>
>>Would the following rearrangement solve the problem?
>>>
>>degC = 5 * (degF - 32) / 9;

>>
>Did you read the answer to the question (<http://c-faq.com/expr/
>truncation1.ht ml>)?

>
Yes I read and it says:
>
"If both operands of a binary operator are integers, C performs an
integer operation,..."

Which is why it does not work.

But in my rewriting there are no binary operation with integer values.

Which is why it would work.

That's why I'm asking if it would work.

You obviously did not make it clear enough to Robert that you had read
it. Don't worry about it.
--
Flash Gordon

[CBFalconer]

Re: Question on the FAQ

Topi Linkala wrote:

>
Question 3.15:
>
Why does the code
double degC, degF;
degC = 5 / 9 * (degF - 32);
keep giving me 0?

Think about it. 5 and 9 are ints. What is the value of 5/9?. How
many times can you subtract 9 from 5 and have the result
non-negative? What is the result of that multiplied by (degF -32)?

>
Would the following rearrangement solve the problem?
degC = 5 * (degF - 32) / 9;

That depends on the type of degF. For double, yes. Much simpler
to use 5/9.0.

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home .att.net>
Try the download section.

** Posted from http://www.teranews.com **

[Ian Collins]

Re: Question on the FAQ

CBFalconer wrote:

Topi Linkala wrote:

>Question 3.15:
>>
>Why does the code
> double degC, degF;
> degC = 5 / 9 * (degF - 32);
>keep giving me 0?

>
Think about it. 5 and 9 are ints. What is the value of 5/9?. How
many times can you subtract 9 from 5 and have the result
non-negative? What is the result of that multiplied by (degF -32)?
>

>Would the following rearrangement solve the problem?
> degC = 5 * (degF - 32) / 9;

>
That depends on the type of degF.

If you read the OP, degF was declared as double.

--
Ian Collins.

[Andrey Tarasevich]

Re: Question on the FAQ

Topi Linkala wrote:

...
But what stops the compiler to calculate the 5/9 first as it is a
constant expression and can be done on compile time?

Nothing really. In the expresison 'E * 5/9' the compiler is free to
calculate '5/9' in advance. What the compiler _cannot_ do is to destroy
the semantics if the full expression (derived from the grammar) and
evaluate '5/9' as an integral division. If the compiler want to
pre-evaluate '5/9', it can do that, but it has to make sure that the
result is precise enough so that the whole thing works "as if" it
evaluated 'E * 5' first and then divided the result by '9'.

AFAIK only guaranteed sequencing on expressions are:
>
1. precedence

Precedence does not guarantee any sequencing. If defines the semantics
of the expression, i.e. it disambiguates the result. It tells you that
'2 + 2 * 2' is 6 and not 8. But the compiler is perfectly free to forget
about these 2's and obtain that final 6 as '37 / 5 + 1' if it is so
inclined.

2. comma operation
3. boolean operations || and &&
4. ?:

These three do indeed introduce real sequencing.

So an expression a*b/c can be calculated (a*b)/c or a*(b/c).

Yes, it can be calculated as 'a*(b/c)' as long as the result is the same
as '(a*b)/c'. If on the given platform 'a*(b/c)' give a different
result, then it the compiler will have no choice but calculate it as
'(a*b)/c'.

--
Best regards,
Andrey Tarasevich