Loading data from file in C - c

i am really struggling to load some numeric floating point data from a file into a C program...the file has floating point numbers with precision of 3 decimal points, each of which is in a single line...i wanted to load these values into an float array and then perform some calculations in it...however i tried loading it into an array of floats but it didn't store the values properly...i.e values like 123.45 were stored as 123.44999 or something..but i don't want this.
i tried to store this in a array of strings and then may be convert them into scaled integers. but i cannot get it to load into an array of string. can anyone tell me where i am going wrong. what i did is something like this.
unsigned char **data
............
data = malloc(sizeof(unsigned char *) * fileSize);
............
while (!feof(fp))
{
if (fscanf (fp, "%s", &data[j]) == 1) // if converted, increment counter
++j;
}
...........
i am a newbie so i am not so good with pointers. and after i load it into an string array, how do i convert it into scaled integers?

You may not have much choice when it comes to floating point numbers like 123.45 being displayed as 123.44999
The 'inaccuracy' of floating point representations has been discussed many times on SO, here's one example.
I would suggest you go back to treating the input as floats, but read up on accuracy issues.

don't use float, use double.
also you can use
double d;
read(d);
(int)(d*100) to get the int.

Unfortunately all you are hitting is the fact that some numbers cannot be accurately represented with a floating point data type. If you manually assigned 124.45 to a float in C (i.e. float f = 123.45) you get exactly the same problem.
Do you really need the level of accuracy? What are you plans for the data?

I would suggest that you use an ifstream to read into a string then convert it to a double.
string number;
ifstream yourfile("yourfile", ios::in);
if(yourfile)
while(yourfile >> number){
double yourFloat=strtod( number.c_str() );
}

How essential is this precision to your application? Floating point numbers can often be slightly inaccurate because they store numbers in a base 2 (binary) representation which cannot properly represent some base-10 numbers.
You have a few options, depending upon how important this precision is to your app.
Live with it. Maybe this inaccuracy isn't a problem for your application
Use Double. It has twice the space
to represent numbers, and will be
more accurate. It can still be inaccurate
in some cases.
Use fixed point math. If you only need to go to n decimal places, it might be worth looking this approach (as it eliminates the source of inaccuracy).
Keep the float as a character string or store it as a Binary Coded Decimal. You'll need to write functions for any mathematical operations you need and performance will be the poorest of these 4 options. But, you'll be able to work with decimal numbers of arbitrary precision without worrying about any loss of precision.
Options 3 or 4 require either the use of an external library, or rolling your own implementation. If 1 or 2 is good enough, then I wouldn't bother with these options.

Related

Trying to recreate printf's behaviour with doubles and given precisions (rounding) and have a question about handling big numbers

I'm trying to recreate printf and I'm currently trying to find a way to handle the conversion specifiers that deal with floats. More specifically: I'm trying to round doubles at a specific decimal place. Now I have the following code:
double ft_round(double value, int precision)
{
long long int power;
long long int result;
power = ft_power(10, precision);
result = (long long int) (value * power);
return ((double)result / power);
}
Which works for relatively small numbers (I haven't quite figured out whether printf compensates for truncation and rounding errors caused by it but that's another story). However, if I try a large number like
-154584942443242549.213565124235
I get -922337203685.4775391 as output, whereas printf itself gives me
-154584942443242560.0000000 (precision for both outputs is 7).
Both aren't exactly the output I was expecting but I'm wondering if you can help me figure out how I can make my idea for rounding work with larger numbers.
My question is basically twofold:
What exactly is happening in this case, both with my code and printf itself, that causes this output? (I'm pretty new to programming, sorry if it's a dumb question)
Do you guys have any tips on how to make my code capable of handling these bigger numbers?
P.S. I know there are libraries and such to do the rounding but I'm looking for a reinventing-the-wheel type of answer here, just FYI!
You can't round to a particular decimal precision with binary floating point arithmetic. It's not just possible. At small magnitudes, the errors are small enough that you can still get the right answer, but in general it doesn't work.
The only way to round a floating point number as decimal is to do all the arithmetic in decimal. Basically you start with the mantissa, converting it to decimal like an integer, then scale it by powers of 2 (the exponent) using decimal arithmetic. The amount of (decimal) precision you need to keep at each step is roughly (just a bit over) the final decimal precision you want. If you want an exact result, though, it's on the order of the base-2 exponent range (i.e. very large).
Typically rather than using base 10, implementations will use a base that's some large power of 10, since it's equivalent to work with but much faster. 1000000000 is a nice base because it fits in 32 bits and lets you treat your decimal representation as an array of 32-bit ints (comparable to how BCD lets you treat decimal representations as arrays of 4-bit nibbles).
My implementation in musl is dense but demonstrates this approach near-optimally and may be informative.
What exactly is happening in this case, both with my code and printf itself, that causes this output?
Overflow. Either ft_power(10, precision) exceeds LLONG_MAX and/or value * power > LLONG_MAX.
Do you guys have any tips on how to make my code capable of handling these bigger numbers?
Set aside various int types to do rounding/truncation. Use FP routines like round(), nearby(), etc.
double ft_round(double value, int precision) {
// Use a re-coded `ft_power()` that computes/returns `double`
double pwr = ft_power(10, precision);
return round(value * pwr)/pwr;
}
As well mentioned in this answer, floating point numbers have binary characteristics as well as finite precision. Using only double will extend the range of acceptable behavior. With extreme precision, the value computed with this code be close yet potentially only near the desired result.
Using temporary wider math will extend the acceptable range.
double ft_round(double value, int precision) {
double pwr = ft_power(10, precision);
return (double) (roundl((long double) value * pwr)/pwr);
}
I haven't quite figured out whether printf compensates for truncation and rounding errors caused by it but that's another story
See Printf width specifier to maintain precision of floating-point value to print FP with enough precision.

How to round 8.475 to 8.48 in C (rounding function that takes into account representation issues)? Reducing probability of issue

I am trying to round 8.475 to 8.48 (to two decimal places in C). The problem is that 8.475 internally is represented as 8.47499999999999964473:
double input_test =8.475;
printf("input tests: %.20f, %.20f \n", input_test, *&input_test);
gives:
input tests: 8.47499999999999964473, 8.47499999999999964473
So, if I had an ideal round function then it would round 8.475=8.4749999... to 8.47. So, internal round function is no appropriate for me. I see that rounding problem arises in cases of "underflow" and therefore I am trying to use the following algorithm:
double MyRound2( double * value) {
double ad;
long long mzr;
double resval;
if ( *value < 0.000000001 )
ad = -0.501;
else
ad = 0.501;
mzr = long long (*value);
resval = *value - mzr;
resval= (long long( resval*100+ad))/100;
return resval;
}
This solves the "underflow" issue and it works well for "overflow" issues as well. The problem is that there are valid values x.xxx99 for which this function incorrectly gives bigger value (because of 0.001 in 0.501). How to solve this issue, how to devise algorithm that can detect floating point representation issue and that can round taking account this issue? Maybe C already has such clever rounding function? Maybe I can select different value for constant ad - such that probability of such rounding errors goes to zero (I mostly work with money values with up to 4 decimal ciphers).
I have read all the popoular articles about floating point representation and I know that there are tricky and unsolvable issues, but my client do not accept such explanation because client can clearly demonstrate that Excel handles (reproduces, rounds and so on) floating point numbers without representation issues.
(The C and C++ standards are intentionally flexible when it comes to the specification of the double type; quite often it is IEEE754 64 bit type. So your observed result is platform-dependent).
You are observing of the pitfalls of using floating point types.
Sadly there isn't an "out-of-the-box" fix for this. (Adding a small constant pre-rounding just pushes the problem to other numbers).
Moral of the story: don't use floating point types for money.
Use a special currency type instead or work in "pence"; using an integral type instead.
By the way, Excel does use an IEEE754 double precision floating point for its number type, but it also has some clever tricks up its sleeve. Essentially it tracks the joke digits carefully and also is clever with its formatting. This is how it can evaluate 1/3 + 1/3 + 1/3 exactly. But even it will get money calculations wrong sometimes.
For financial calculations, it is better to work in base-10 to avoid represenatation issues when going to/from binary. In many countries, financial software is even legally required to do so. Here is one library for IEEE 754R Decimal Floating-Point Arithmetic, have not tried it myself:
http://www.netlib.org/misc/intel/
Also note that working in decimal floating-point instead of fixed-point representation allows clever algoritms like the Kahan summation algorithm, to avoid accumulation of rounding errors. A noteworthy difference to normal floating point is that numbers with few significant digits are not normalized, so you can have e.g both 1*10^2 and .1*10^3.
An implementation note is that one representation in the std uses a binary significand, to allow sw implementations using a standard binary ALU.
How about this one: Define some threshold. This threshold is the distance to the next multiple of 0.005 at which you assume that this distance could be an error of imprecision. Execute appropriate methods if it's within that distance and smaller. Round as usual and at the end, if you detected that it was, add 0.01.
That said, this is only a work around and somewhat of a code smell. If you don't need too much speed, go for some other type than float. Like your own type that works like
class myDecimal{ int digits; int exponent_of_ten; } with value = digits * E exponent_of_ten
I am not trying to argument that using floating point numbers to represent money is advisable - it is not! but sometimes you have no choice... We do kind of work with money (life incurance calculations) and are forced to use floating point numbers for everything including values representing money.
Now there are quite some different rounding behaviours out there: round up, round down, round half up, round half down, round half even, maybe more. It looks like you were after round half up method.
Our round-half-up function - here translated from Java - looks like this:
#include <iostream>
#include <cmath>
#include <cfloat>
using namespace std;
int main()
{
double value = 8.47499999999999964473;
double result = value * pow(10, 2);
result = nextafter(result + (result > 0.0 ? 1e-8 : -1e-8), DBL_MAX);
double integral = floor(result);
double fraction = result - integral;
if (fraction >= 0.5) {
result = ceil(result);
} else {
result = integral;
}
result /= pow(10, 2);
cout << result << endl;
return 0;
}
where nextafter is a function returning the next floating point value after the given value - this code is proved to work using C++11 (AFAIK the nextafter is also available in boost), the result written into the standard output is 8.48.

c: change variable type without casting

I'm changing an uint32_t to a float but without changing the actual bits.
Just to be sure: I don't wan't to cast it. So float f = (float) i is the exact opposite of what I wan't to do because it changes bits.
I'm going to use this to convert my (pseudo) random numbers to float without doing unneeded math.
What I'm currently doing and what is already working is this:
float random_float( uint64_t seed ) {
// Generate random and change bit format to ieee
uint32_t asInt = (random_int( seed ) & 0x7FFFFF) | (0x7E000000>>1);
// Make it a float
return *(float*)(void*)&asInt; // <-- pretty ugly and nees a variable
}
The Question: Now I'd like to get rid of the asInt variable and I'd like to know if there is a better / not so ugly way then getting the address of this variable, casting it twice and dereferencing it again?
You could try union - as long as you make sure the types are identical in memory sizes:
union convertor {
int asInt;
float asFloat;
};
Then you can assign your int to asFloat (or the other way around if you want to). I use it a lot when I need to do bitwise operations on one hand and still get a uint32_t representation on the number on the other hand
[EDIT]
Like many of the commentators rightfully state, you must take into consideration values that are not presentable by integers like like NAN, +INF, -INF, +0, -0.
So you seem to want to generate floating point numbers between 0.5 and 1.0 judging from your code.
Assuming that your microcontroller has a standard C library with floating point support, you can do this all standards compliant without actually involving any floating point operations, all you need is the ldexp function that itself doesn't actually do any floating point math.
This would look something like this:
return ldexpf((1 << 23) + random_thing_smaller_than_23_bits(), -24);
The trick here is that we happen to know that IEEE754 binary32 floating point numbers have integer precision between 2^23 and 2^24 (I could be off-by-one here, double check please, I'm translating this from some work I've done on doubles). So the compiler should know how to convert that number to a float trivially. Then ldexp multiplies that number by 2^-24 by just changing the bits in the exponent. No actual floating point operations involved and no undefined behavior, the code is fully portable to any standard C implementation with IEEE754 numbers. Double check the generated code, but a good compiler and c library should not use any floating point instructions here.
If you want to peek at some experiments I've done around generating random floating point numbers you can peek at this github repo. It's all about doubles, but should be trivially translatable to floats.
Reinterpreting the binary representation of an int to a float would result in major problems:
There are a lot of undefined codes in the binary representation of a float.
Other codes represent special conditions, like NAN, +INF, -INF, +0, -0 (sic!), etc.
Also, if that is a random value, even if catching all non-value representations, that would yield a very bad random distribution.
If you are working on an MCU without FPU, you should better think about avoiding float at all. An alternative might be fraction or scaled integers. There are many implementations of algorithms which use float, but can be easily converted to fixed point types with acceptable loss of precision (or even none at all). Some might even yield more precision than float (note that single precision float has only 23 bits of mantissa, an int32 would have 31 bits (+ 1 sign for either), same for a fractional or fixed scaled int.
Note that C11 added (optional) support for _Frac. You might want to research on that.
Edit:
According you your comments, you seem to convert the int to a float in range 0..<1. For that, you can assemble the float using bit operations on an uint32_t (e.g. the original value). You just need to follow the IEEE format (presumed your toolchain does comply to the C standard! See wikipedia.
The result (still uint32_t) can then be reinterpreted by a union or pointer as described by others already. Pack that in a system-dependent, well-commented library and dig it deep. Do not forget to check about endianess and alignment (likely both the same for float and uint32_t, but important for the bit-ops).

Can't round number to two decimal places using floorf

I found this formula here:
podrucje[i][j] = floorf(podrucje[i][j] * 100 + 0.5)/100;
where podrucje[][] is float matrix. floorf returns correct value (desired * 100) but problem is with /100. For example if I have 49.599998, floorf returns 4960.000 and after dividing by 100 result is again 49.599998.
Where is the problem?
I know I can round it while printing but I'm going to use this matrix in Excel where it is converted to range thus i can't influence on representation.
Most decimal numbers aren't exactly representable as floats.
You probably need to either store them as integers, or do the rounding when converting to something humanly visible (i.e. printing the number out).
If you want to use the number for further calculations and need it to be rounded first, use a different representation.
This is a fundamental problem with the representation of floating point arithmetic which is not base-10! You can print a base-10 with rounding to 2 decimal places though.
The standard safe way to export numbers safely across boundaries is through plain text ascii rather than as binary.

How do I convert the exponential formats to the decimal format in c

I would like to convert the exponential number 5.52794e+15.
It won't fit in an int, but a long long will do:
long long x = 5.52794e+15;
Alternative: you want to print it as an integer.
double d = 5.52794e+15;
printf("%15.0lf\n",d );
Gives:
5527940000000000
Besides using long longs, you could also:
Use a packed decimal library.
You could use logarithmic (or geometric, etc) scaling in a standard int or long.
Use a structure of ints, representing the significand and exponent, and do floating point math manually (or with a good library).
I've used all three methods, they all have their ups and downs. Packed Decimal is slowest and most accurate. Logarithmic scaling is by far the fastest and easiest to implement, and least accurate. Reproducing floating point via integers is in-between in performance, essentially the same in accuracy as "real" floating point, and hardest to implement.
All 3 are slower than using floating point hardware- assuming your hardware has floating point!
long long x = 5.52794e+15;
where the exponent "15" has to be an integer, and the coefficient 5.52794 is any real number or could even be integer.
The maximum value of the mantissa is just over 32,000,000 Any value above this will be truncated.
This becomes significant where you have a running total and are adding in small values and keeping an accurate total. - for example power usage where the value that you are adding in is not a whole number.

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