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What is meant by . usage after a number in Fortran?
(1 answer)
Closed 3 years ago.
I have the following function signature in Fortran. I am new to the Fortran language - what is meant by 0. in the function below?
ANTY(IAN,NA,0.,0.)
The notation 0. with a decimal point represents a floating point number, while 0 without a decimal point represents an integer. This is an important distinction for some implementations of FORTRAN, where the programmer must ensure that data types are consistent between the caller of a subroutine and the subroutine itself.
It is equivalent to 0.0 as a 32 bit real number (float in C)
Related
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What is the behavior of integer division?
(6 answers)
Closed 6 years ago.
So I'm trying my hand at C++ again and I'm writing a function that keeps giving the wrong value.
I've boiled it down to a line basically this: double testD = (10/100); that gives me 0 instead of 0.1.
I might have missed something major but I thought double held decimal values? For the life of me I can't get this to be correct.
Please help me (even if I'm terribly stupid).
10 and 100 are integers. This will result in integer division, which will evaluate to 0. That 0 will then be stored in testD.
To force floating-point division cast one of the numbers to float or double.
((double)whatever)/something
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Floating point arithmetic not producing exact results [duplicate]
(7 answers)
Closed 8 years ago.
In C++, floor(9099.96 *100.0) is giving me the answer as 909995. I am expecting 909996.
I am not able to think of explanations here. Any help will be appreciated.Thanks.
This is the proper result: according to IEEE754 calculator, the value of 9099.96 is represented as 9099.9599609375 in double. After multiplication by 100 you get 909995.99609375. Taking floor gives you 909995.
Try ceil. It will always round the the next greatest number, which in this case will give you 909996. You may also want to look at the round function.
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Using extremely large integer holding 3001 digits
(2 answers)
Closed 9 years ago.
How can i handle very large integers like 2^100000000 in c++?
I found no solution for this on internet that gives an exact answer.
Is there any mechanism that gives correct value in c++ for such large integers?
What you are looking for is called arbitrary precision arithmetic, you will find numerous libraries and educational resources with some googling.
You can represent given number as a string and convert it to array with integer digits. But the simplest way to google by keywords "long arithmetic c++ library" or something.
Maybe you want to use a Computer Algebra System (CAS), which would represent your expression like this:
class Pow : public Expr {
Number base;
Number exp;
};
Pow expr = new Pow(2, 100*1000*1000);
A CAS then allows you to manipulate these expressions structurally instead of the concrete values.
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Why does floating-point arithmetic not give exact results when adding decimal fractions?
(31 answers)
Closed 6 years ago.
I've been trying to use Fortran for my research project, with the GNU Fortran compiler (gfortran), latest version,
but I've been encountering some problems in the way it processes real numbers. If you have for example the code:
program test
implicit none
real :: y = 23.234, z
z = y * 100000
write(*,*) y, z
end program
You'll get as output:
23.23999 2323400.0
I find this really strange.
Can someone tell me what's exactly happening here? Looking at z I can see that y does retain its precision, so for calculations that shouldn't be a problem I suppose. But why is the output of y not exactly the same as the value that I've specified, and what can I do to make it exactly the same?
This is not a problem - all you see is floating-point representation of the number in the computer. The computer cannot handle real numbers exactly, but only approximations of them. A good read about this can be found here: What Every Computer Scientist Should Know About Floating-Point Arithmetic.
Simply by replacing real with double precision, you can increase the number of significant decimal places from about six to about 15 on most platforms.
The general issue is not limited to Fortran, but the representation of base 10 real numbers in another base of finite precision. This computer science question is asked many times here.
For the specifically Fortran aspects, the declaration "real" will likely give you a single precision floating point. As will expressing a constant as "23.234" without a type qualifier. The constant "100000" without a decimal point is an integer so the expression "y * 100000" is causing an implicit conversion of an integer to a real because "y" is a real variable.
For previous some previous discussions of these issues see Extended double precision , Fortran: integer*4 vs integer(4) vs integer(kind=4) and Is There a Better Double-Precision Assignment in Fortran 90?
The problem here is not with Fortran, in fact it is not a problem at all. This is just a feature of floating-point arithmetic. If you think about how you would represent 23.234 as a 'single float' in binary, you would see that the number has to be saved to only so many decimals of precision.
The thing to remember about float point number is: numbers that look round and even in base-10 probably won't in binary.
For a brief overview of floating-point topics, check the Wikipedia article. And for a VERY thorough explanation, check out the canonical paper by Goldberg (PDF).
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Checking if a double (or float) is NaN in C++
(21 answers)
Closed 1 year ago.
Is it possible to check if a number is NaN or not?
Yes, by use of the fact that a NaN is not equal to any other number, including itself.
That makes sense when you think about what NaN means, the fact that you've created a value that isn't really within your power to represent with "normal" floating point values.
So, if you create two numbers where you don't know what they are, you can hardly consider them equal. They may be but, given the rather large possibility of numbers that it may be (infinite in fact), the chances that two are the same number are vanishingly small :-)
You can either look for a function (macro actually) like isnan (in math.h for C and cmath for C++) or just use the property that a NaN value is not equal to itself with something like:
if (myFloat != myFloat) { ... }
If, for some bizarre reason, your C implementation has no isnan (it should, since the standard mandates it), you can code your own, something like:
int isnan_float (float f) { return (f != f); }
Under Linux/gcc, there's isnan(double), conforming to BSD4.3.
C99 provides fpclassify(x) and isnan(x).
(But C++ standards/compilers don't necessarily include C99 functionality.)
There ought to be some way with std::numeric_limit<>... Checking...
Doh. I should have known... This question has been answered before...
Checking if a double (or float) is NaN in C++
Using NaN in C++?
http://bytes.com/topic/c/answers/588254-how-check-double-inf-nan
you are looking for null, but that is only useful for pointers. a number can't be null itself, it either has a known value that you put in there or random data from whatever was there in memory before.