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# type conversion

Posted on 2004-09-03
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Hi,
I was recently debugging a program, when I noticed something funny going on with the maths - with some playing, I managed to determine that this was to do with type conversions - trouble is, I can't figure out exactly *why* these 'problems' are there! Could somebody please explain this to me? I'm using VC++ 5.0. Take for example the following situation:

float a=4.7, b=50;

float c=a*b;                         //c=235, correct.
int d=a*b;                           //d=234!
int e=(float)4.7*(float)50;     //e=235 :S

What is causing this?

Thanks alot,
Jonathan
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Question by:jonathandodd

LVL 11

Expert Comment

ID: 11978198
you have initialised b with an integer not a float number you have to do

b=50.0
0

Author Comment

ID: 11978212
that won't make any difference...if it's a float variable it'll be stored as a float, extra .00 s  or not :p Anyhow, that was just a setup scenario to demonstrate the problem...
0

LVL 11

Accepted Solution

avizit earned 85 total points
ID: 11978282
the reason i can think of is

in

int d=a*b;

teh product is getting calculated to   234.99999999999999 ( something like that ) and the
conversion to int is making it 234

while in

int e=(float)4.7*(float)50;
the product is getting calclated to

235 or maybe 235.00000000000000000001 ..

also check http://www2.hursley.ibm.com/decimal/decifaq1.html#inexact

0

LVL 22

Expert Comment

ID: 11978285
what's happening is normal floating point roundoff.

Apparently 4.7 times 50.000 is a few bits shy of 235.

If you print out the number with lots of decimal places you'll see something like 234.999998.

Storing the result into an integer results in truncation to the next lower integer.

0

LVL 3

Expert Comment

ID: 11978332
The real question question is why (float)4.7*(float)50 is different from a*b when a and b have type float.  The real answer is that in the case of (float)4.7*(float)50, the compiler is doing the math at compile time, and in the case of a*b, the code generated by the compiler is doing the math at runtime.  Your compiler seems to use higher-precision math than float while preprocessing literals, and it's allowed to do that.  It's not allowed to inject code to do the same thing into the final executable because the programmer specified float*float and nothing else.  Note also that the preprocessor probably ignores your casts, so 4.7*50 might work the same way as (float)4.7*(float)50.
0

Author Comment

ID: 11979358
wow thanks - I've been working with C++ for years and never realised floating point calculations worked like that! Why is it that floating point calculations are sometimes worked out to be 0.000000001 or something off? Surely, if I put in 4.7, it stores it as 4.7, not 4.699999999999999 or 4.7000000000000001 (or to however many places)?
0

Author Comment

ID: 11980069
OK I've done some reading - I geddit now :D

thanks again!
Jonathan
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