Rounding down/truncating double - c++

I'm doing calculations with financial data formatted as follows:
<up to 5 digits>.<two digits>
Basically, in my program I'm encountering a floating point error. For example, if I have:
11.09 - (11.09 * 0.005) = 11.03455
I want to be able to use 11.03455 and not what's generated: 11.0345499999999...
I'm comparing values that my program generates with values I have in text files that are in string format. I only need two decimal points of precision and I can round down. Is there a way that I can cut this to 11.03?
I was thinking it would be easiest if I turn this into a string and just parse it character by character, only adding two characters past the '.' character. Is this a good solution? Any better ideas?
Here is what I have:
string dataProcessor::createTwoDec(double price){
string s = to_string(price);
string output = "";
int dist = 0;
int num_wanted = 0;
bool pt_found = false;
for(int i = 0; i < s.length(); i++){
if(s[i] == '.')
pt_found = true;
if(pt_found)
dist++;
if(dist > 3)
break;
output += s[i];
num_wanted++;
}
return output.substr(0, num_wanted);
}

You can use the following formula for round-off by n decimal places (n is not too large):
round(x*10^n)/10^n
where n is number of decimal places required.
In your case, n is 5. Hence, it will be
result = round(result*100000)/100000;
See How do you round off decimal places in C++?

Related

CodeSignal isLucky task c++

Started to practice C++ by trying to do some tasks in CodeSignal, but I can’t figure it out why it has an output which is always false. Saw a similar answer to this task but didn’t want to copy and paste without understanding where the error is.
Ticket numbers usually consist of an even number of digits. A ticket number is considered lucky if the sum of the first half of the digits is equal to the sum of the second half.
Example:
For n = 1230, the output should be isLucky(n) = true
For n = 239017, the output should be isLucky(n) = false
Code:
bool isLucky(int n) {
string convert = to_string(n); // to convert from string to int
int sizehalbe = convert.size() / 2; //divide into 2 halfs
//Stor each half
string h1 = convert.substr(0, sizehalbe-1);
string h2 = convert.substr(sizehalbe, convert.size()-1);
int sum1=0, sum2=0; //Calculate the halfs
for(int i=0;i<h1.length();i++)
{
sum1 += int(h1.at(i));
}
for(int j=0;j<h2.length();j++)
{
sum2 += int(h2.at(j));
}
if(sum1 == sum2)
return true;
else
return false;
}
(1). Foremost your h1 always miss one digit so instead of
h1 = convert.substr(0, sizehalbe-1);
that's the only main issue your code has, convert should be gone till sizehalbe
string h1 = convert.substr(0, sizehalbe);
(2). whenever you typecast from character to integer, check what it gives
cout<<int('0'); will give you 48 instead of 0.
in particular, this case it's not changed your main output
(due to both sum1 & sum2 will get higher result than what actually should be,
but get same level of higher.)
sum½ += int(h½.at(i)) - 48;
(3). you can optimize your last condition.
when boolean result is depending on condition you can do
return (sum1 == sum2);

I need to get a random number in the range from a to b with n decimal places

I need a function that returns me a random number with n decimal places
Example:
int aleatorio(int li, int ls)
{
return rand()%(ls+1-li)+li;
}
What i want is:
float new_random(int start, int final, int number_decimals)
{
return // What should I write here?
}
if I would call this function 5 times like this::
new_random(0, 5, 4);
The exit would be:
0.2344
3.4356
2.8435
4.2435
I do not want to use this, because I need numbers of 4 exact decimal places since I will not use them to print, but you will have others:
cout << setprecision(4) << 4.24359675967 << endl; //I do not want this
I need numbers of 4 exact decimal places
Then you cannot use finite precision binary floating point (i.e. float, double or long double) because those types cannot exactly represent all of the values with 4 decimal places.
A solution is to use arbitrary precision floating point, and another is to use fixed point. C++ standard doesn't provide arbitrary precision types nor fixed point types. Another approach is to give up the requirement of exactly representing those values and accept the almost exact values that are achievable with limited precision.
Try this for a Java solution. Multiply the start and finish by 1000, generating ints between the range and then divide the resultant number by 1000 as a double.
int start = 20;
int finish = 30;
int count = 10;
Random r = new Random();
r.ints(start * 1000, finish * 1000).filter(n -> n % 10 != 0)
.limit(count).mapToDouble(n -> n / 1000.)
.forEach(System.out::println);
prints something like this.
21.186
26.983
25.345
20.764
27.911
21.139
24.679
27.722
29.443
28.675
Or as a method supplying the starting number, ending number and precision.
for (int i = 0; i < 10; i++) {
System.out.println(newRandom(start,finish,4));
}
static Random r = new Random();
public static Double newRandom(int start, int finish, int precision) {
int f = (int)Math.pow(10,precision);
return r.ints(start * f, finish * f).filter(n -> n % 10 != 0)
.limit(1).mapToDouble(n -> n / (double)f).findFirst().getAsDouble();
}
Prints something like this.
28.4444
25.0259
29.5611
25.6445
25.4977
28.5124
28.9709
23.4835
27.9766
23.9438
You can generate an integer number N between start and final * 10^number_decimals and then return N / 10^number_decimals
Eg. start = 0, final = 5, number_decimals = 4 ==> N in [0 - 50000] ==> N/10000 in [0.0000 - 5.0000]
float new_random(int start, int final, int number_decimals) {
return aleatorio(start, final*pow10(number_decimals))/number_decimals;
}
You can define pow10 as:
int pow10(int p) {
if (p == 0) return 1;
else return 10 * pow10(p-1);
}

my run-length encoding doesn't work with big numbers

I have a assingment were I need to code and decode txt files, for example: hello how are you? has to be coded as hel2o how are you? and aaaaaaaaaajkle as a10jkle.
while ( ! invoer.eof ( ) ) {
if (kar >= '0' && kar <= '9') {
counter = kar-48;
while (counter > 1){
uitvoer.put(vorigeKar);
counter--;
}
}else if (kar == '/'){
kar = invoer.get();
uitvoer.put(kar);
}else{
uitvoer.put(kar);
}
vorigeKar = kar;
kar = invoer.get ( );
}
but the problem I have is if need to decode a12bhr, the answer is aaaaaaaaaaaabhr but I can't seem to get the 12 as number without problems, I also can't use any strings or array's.
c++
I believe that you are making following mistake: imagine you give a32, then you read the character a and save it as vorigeKar (previous character, I am , Flemish so I understand Dutch :-) ).
Then you read 3, you understand that it is a number and you repeat vorigeKar three times, which leads to aaa. Then you read 2 and repeat vorigeKar two times, leading to aaaaa (five times, five equals 3 + 2).
You need to learn how to keep on reading numeric characters, and translate them into complete numbers (like 32, or 12 in your case).
Like #Dominique said in his answers, You're doing it wrong.
Let me tell you my logic, you can try it.
Pesudo Code + Logic:
Store word as a char array or string, so that it'll be easy to print at last
Loop{
Read - a //check if it's number by subtracting from '0'
Read - 1 //check if number = true. Store it in int res[] = res*10 + 1
//Also store the previous index in an index array(ie) index of char 'a' if you encounter a number first time.
Read - 2 //check if number = true. Store it in res = res*10 + 2
Read - b , h and so on till "space" character
If you encounter another number, then store it's previous character's index in index array and then store the number in a res[] array.
Now using index array you can get the index of your repeating character to be printed and print it for it's corresponding times which we have stored in the result array.
This goes for the second, third...etc:- numbers in your word till the end of the word
}
First, even though you say you can't use strings, you still need to know the basic principle behind how to turn a stream of digit characters into an integer.
Assuming the number is positive, here is a simple function that turns a series of digits into a number:
#include <iostream>
#include <cctype>
int runningTotal(char ch, int lastNum)
{
return lastNum * 10 + (ch -'0');
}
int main()
{
// As a test
char s[] = "a123b23cd1/";
int totalNumber = 0;
for (size_t i = 0; s[i] != '/'; ++i)
{
char digit = s[i]; // This is the character "read from the file"
if ( isdigit( digit) )
totalNumber = runningTotal(digit, totalNumber);
else
{
if ( totalNumber > 0 )
std::cout << totalNumber << "\n";
totalNumber = 0;
}
}
std::cout << totalNumber;
}
Output:
123
23
1
So what was done? The character array is the "file". I then loop for each character, building up the number. The runningTotal is a function that builds the integer from each digit character encountered. When a non-digit is found, we output that number and start the total from 0 again.
The code does not save the letter to "multiply" -- I leave that to you as homework. But the code above illustrates how to take digits and create the number from them. For using a file, you would simply replace the for loop with the reading of each character from the file.

C/C++ - "Negative" string converted to zero

I have the following .txt file:
{{1,2,3,0}, {1,1,1,2}, {0,−1,3,9}}
This is a 3x4 matrix. I'm using strtok to extract the numbers and saving on a float matrix. The problem is, when p gets -1, it's being converted to zero when saved on matrix. How could I fix it?
p = strtok(&matrix[0u], " {},");
for (i = 0; i < m + 1; i++){
for (j = 0; j < n + 1; j++) {
aux[i][j] = atoi(p);
if (p)
p = strtok(NULL, " {},");
}
}
Is there a better way to extract the numbers, one at a time? How?
Your minus sign doesn't work. Compare:
this - is the ASCII minus sign
this − is your character whixh might be called "minus sign" by Unicode, but it is not normally recognised as such by C++ library functions
Don't copy code from Word documents and like places. If in doubt, convert to ASCII with iconv or a similar utility.

Why does trunc(1) output as 0?

Can someone explain me why in c++ happens such a thing:
double tmp;
... // I do some operations with tmp
// after which it has to be equal to one
cout << tmp; // prints 1
cout << trunc(tmp); // prints 0
cout << trunc(tmp*10); // prints 9
I am using this for separation part right of decimal part from the number for example if i have: 5.010 ... i want to have 0.010 .. so I am using:
double remainder = tmp - trunc(tmp);
I am posting the whole code....the suggestion with floor does not worked
short getPrecision(double num, short maxPrecision) {
// Retrieve only part right of decimal point
double tmp = fabs(num - trunc(num));
double remainder = tmp;
// Count number of decimal places
int c = 0;
while (remainder > 0 && c < maxPrecision) {
tmp *= 10;
remainder = tmp - trunc(tmp);
c++;
}
return c;
}
When I run this function for example with 5.1 the remanider is 0 instead of 1
After some calculations it has to be one? Well, it could as well be 0.99999999999999999. Floating point operations are not precise, you should always take that into account.
Please see picture at http://en.cppreference.com/w/cpp/numeric/math/trunc. The chart there explains the inconsistency with truncing 1. Probably the same applies to 10 as well
This should help you achieving what you need:
double remainder = tmp - floor(tmp);