whats going on with the following code:

#include <stdio.h>
void main(void){
        float i;
        i=0.1;
        for(i;i<1000;i++){
                printf("%f", i);
        }

(it will output something like: 0.1, 1.1, 1.1, 1.19999!!! )

note 1:after many for's, a significant difference will be
noticed....this happend with me when I was doing
a program to solve integrals through computational methods...

note 2: putting the variable in my compiler IDE watch window the data seems
to be ok...only when outputting it to the screen using the printf() function
the modified data is shown...unless I ask to printf to use only 2 or three
decimal  places for float (%.2f)

thats about it.. :-)

Henrique Seganfredo

--

14.1:   When I set a float variable to, say, 3.1, why is printf()
        printing it as 3.0999999?

A:      Most computers use base 2 for floating-point numbers as well as
        for integers.  In base 2, 1/1010 (that is, 1/10 decimal) is an
        infinitely-repeating fraction: its binary representation is
        0.0001100110011... .  Depending on how carefully your compiler's
        binary/decimal conversion routines (such as those used by
        printf) have been written, you may see discrepancies when
        numbers (especially low-precision floats) not exactly
        representable in base 2 are assigned or read in and then printed
        (i.e. converted from base 10 to base 2 and back again).  See
        also question 14.6.

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If this is really what you wrote your compiler is broken because there
is no increment operator defined for floats.

Francis Glassborow      Journal Editor, Association of C & C++ Users
64 Southfield Rd
Oxford OX4 1PA          +44(0)1865 246490
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#>#include <stdio.h>
#>void main(void){
#>        float i;
#>        i=0.1;
#>        for(i;i<1000;i++){

# If this is really what you wrote your compiler is broken because there
# is no increment operator defined for floats.

Hmm. ISO 9899 6.3.2.4 says postfix ++ wants a modifiable scalar type.
I'd say a float like the above is a perfect example. Hey, ++ even works
for pointers, try it one day ;-) :-) :-O

Maybe you mixed up ++ with % ?

Regards,

        Jens
--
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SIGSIG -- signature too long (core dumped)
--

Are you sure?

According to H&S p-200 "the operands [of ++ and --] may be of any
scalar type" and p-110 Table 5-1 shows float as a scalar type.  FWIW,
the AIX C compiler handles this with no error.

--

--

On Wed, 01 Sep 1999 06:15:02 GMT, Francis Glassborow

I thought ++ and -- worked for all scalar types, not just integral
ones?

Regards,

                          -=Dave
Just my (10-010) cents
I can barely speak for myself, so I certainly can't speak for B-Tree.
Change is inevitable.  Progress is not.
--

*is amazed*

Never knew that before. Had a look in the FAQ there's nothing about it
there, and according to K&R2, the only restriction is that the operator
must be applied to an lvalue (variable).

variables; an expression like (i+j)++ is illegal."

I'd have thought that they'd have said 'integer variables' there if
inc/dec on floats was illegal.

A similar implication is made on p.203, where the only listed
requirement for operators ++ and -- is that they are applied to an
lvalue.

I realise that K&R is not the standard, and that Appendix A in the back
is merely a description of the standard, but I'd have thought that
that's quite an omission. Almost worth an entry on the 'errata' page.

I also realise that the code a compiler needs to generate to increment a
float would not nearly be as efficient on many architectures as that to
increment an integer, due to the (universal?) integer increment
instruction being more prevalent than the (non-existent?) float
increment instruction, but making the language increment and decrement
operators meaningless when applied to floats does seem a little -
unsymmetric.

Truly surprised,
Adam Spragg.

--
Apparently [...] police in many lands are now complaining that local
arrestees are insisting on having their Miranda rights read to them,
just like perps in American TV cop shows. When it's explained to them
that they are in a different country, where those rights do not exist,
they become outraged. Starsky and Hutch reruns, dubbed into diverse
languages, may turn out, in the long run, to be a greater force for
human rights than the [United States] Declaration of Independence.
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http://www.io.com/~mccoy/beginning_print.html)

----------------
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On Tue, 31 Aug 1999 20:11:56 GMT, "Henrique Seganfredo"

Let me suggest you read ftp://ftp.quitt.net/outgoing/goldberg1.pdf to
learn about floating point numbers.

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 _
Kevin D Quitt  USA 91351-4454           96.37% of all statistics are made up
Per the FCA, this email address may not be added to any commercial mail list
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  ^^^^
How could you do this?  As a good usenet citizen, you have lurked in the
newsgroup and checked the FAQ before posting.  You KNOW this is wrong.

5 is not a power of 2.

As a good usenet citizen, you have already checked the backtaffic on the
newsgroup with a tool like dejanews, so you already know the answer.

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The increment operator (I thought) only incremented integers and not floats.
What would (should) it mean to increment a float?  Should it increase by 1?
0.1? 0.01?  I would expect the answer should be 1.0, though I'm not positive.
For it to be anything else would lead to certain confusion.
--
Open Source middleware available at http://pweb.netcom.com/~tgagne
--

   Nothing strange at all; remember that most real numbers can't be
represented in binary _exactly_. Floating point representation is
nearly always an approximation. In your specific case the problem is
aggravated because you are using float and not more precise double.

--
        Regards,
                Alex Krol
Disclaimer: I'm not speaking for Scitex
Corporation Ltd

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It's good that you didn't know this untruth.

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Thanks. Curiously that was a hole in my thinking.  I would never even
dream of using increment and decrement operators with floating point
types because, from my perspective, they serve no useful purpose.
Integral and pointer types are a different issue (and in both these
cases the operators often map to a machine code instruction).

Actually this highlights a common problem, if we _know_ something is
_useless_ we never consider it until our ignorance hits us.

Francis Glassborow      Journal Editor, Association of C & C++ Users
64 Southfield Rd
Oxford OX4 1PA          +44(0)1865 246490
All opinions are mine and do not represent those of any organisation
--

Reference my earlier ignorance.  The above claim is what compounded my
ignorance.  According to my belated reading of 9899:1990 the output
should be:
0.1, 1.1, 2.1, 3.1, etc.

Or am I still wrong?

True, but it does not seem to relate to the sequence he claimed was
produced.

Francis Glassborow      Journal Editor, Association of C & C++ Users
64 Southfield Rd
Oxford OX4 1PA          +44(0)1865 246490
All opinions are mine and do not represent those of any organisation
--