So if I am understanding c++ and logarithms correctly. Something like this should give me the base that I am looking for?
I am having some issues, but think that this way is a correct start.
#include <iostream>
using namespace std;
int main(void)
{
int x = 2;
int y = 64;
int value = log x (y);
cout << value << endl;
return 0;
}
This should display "6", but I am not sure how to really use the logarithm library..
There are three parts of a logarithm problem. The base, the argument (also called power) and the answer. You are trying to find the number of times 2 (the base) must be multiplied to get 64 (the answer).
So instead of dealing with powers and guessing. Let's just count the number of times we divide the answer 64, by the base, until we get 1.
64 / 2 = 32
32 / 2 = 16
16 / 2 = 8
8 / 2 = 4
4 / 2 = 2
2 / 2 = 1
Here we count 6 lines, so you divided the answer 64, by the base 2, 6 times. This is the same as saying you raised 2 to the 6th power to get 64.
Hmm for that to work you would need the math.h library. If you want to do this without it. You could do a loop where you constantly divide by the base given by the user.
int base = 0;
int answer = 0;
int exponent = 0;
cout << "Please type in your base";
cin >> base;
cout << "Please type in your answer";
cin >> answer;
while (answer != 1)
{
answer /= base;
exponent++
}
cout << "The exponent is " << exponent;
The implementation of a logarithm is most times a tayler function with approximates logarithm to the base e. To get the logarithm to any other base you do something like this:
#include <cmath>
double y = std::log(x)/std::log(base);
Description:
std::log() natural logarithm to base e.
y = result
base = base of the logarithm. Like 2 or 10.
For your reference:
http://en.cppreference.com/w/cpp/numeric/math/log
http://en.cppreference.com/w/c/numeric/math/log
https://en.wikipedia.org/wiki/Logarithm
https://en.wikipedia.org/wiki/Logarithm#Change_of_base
https://en.wikipedia.org/wiki/Taylor_series
https://en.wikipedia.org/wiki/Taylor_series#Examples
Here's a better implementation without using the math.h library. I might have some superfluous code as it's been like a year since I have written in C++. (Shame on me) Thanks for your question it got me wanting to brush up on my C++!
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
float base = 0;
float answer = 0;
int exponent = 0;
cout.precision(4);
cout << "Please type in your base \n";
cin >> base;
cout << "Please type in your answer\n";
cin >> answer;
cout << fixed;
while (answer > base)
{
answer /= base;
exponent++;
}
cout << "The exponent is: " << exponent << endl;
cout << "The Remainder is: " << fixed << answer << endl;
return 0;
}
Related
my goal is to turn a double's fraction part into an integer, for example: turn 0.854 into 854 or turn 0.9321 into 9321 (the whole part of the double is always 0)
i was doing it like this, but i dont know how to fill the while's parameters to make it stop when variable x becomes a whole number without a fraction:
double x;
cin >> x;
while(WHAT DO I TYPE HERE TO MAKE IT STOP WHEN X BECOMES A WHOLE NUMBER){
x *= 10;
}
it would be best if i could do it with a loop, but if it is not possible, i am open to other suggestions :)
That question has no answer in a mathematical/logical sense. You have to read how floating point numbers in computers work, see e.g.
https://en.wikipedia.org/wiki/Floating-point_arithmetic
and understand that they are not decimal point numbers in memory. A floating point number in memory consists of three actual numbers: significant * base^{exponent} and according to IEEE the base used is "2" in basically any modern floating point data, but in even more generality, the base can be anything. Thus, whatever you have in your mind, or even see on your screen as output, is a misleading representation of the data in memory. Your question is, thus, mainly a misconception of how floating point numbers in computers work...
Thus, what you specifically ask for does in general not exist and cannot be done.
However, there may be special application for output formatting or whatever where something like this may make sense -- but then the specific goal must be clearly stated in the question here. And in some of such cases, using a "string-based" approach, as you suggested, will work. But this is not an answer to your generic question and has the high potential to also mislead others in the future.
Actually, one way to make your question obvious and clear is to also specify a fixed desired precision, thus, numbers after the decimal point. Then the answer is quite trivially and correctly:
long int value = fraction * pow(10, precision);
In this scenario you know 100% what your are doing. And if you really like you could subsequently remove zeros from the right side...
int nTrailingZeros = 0;
while (value%10 == 0) {
value /= 10;
++nTrailingZeros;
}
However there is another principle problem on a numerical level: there is no mathematical difference between, e.g., 000003 and just 3, thus in any such application the input 0.000003 will give the same results as 0.0003 or 0.3 etc. This cannot be a desired functionality... it is pretty useless to ask about the *value of the fractional part of a floating point number. But, since we have a known precision`, we can do:
cout << setw(precision-ntrailing) << setfill('0') << value << endl;
which will fill in the eventually missing leading zeros.
See this complete tested test code
#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
int main() {
double v = 0.0454243252;
int precision = 14;
long int value = v*pow(10,precision);
cout << value << endl;
// 4542432520000
int nTrailingZeros = 0;
while (value%10 == 0) {
value /= 10;
++nTrailingZeros;
}
cout << value << endl;
// 454243252
cout << setw(precision-ntrailing) << setfill('0') << value << endl;
// 0454243252
}
Here is a possible approach:
#include <iostream>
#include <string>
#include <sstream>
#include <iomanip>
using namespace std;
int main()
{
cout << "Enter Number: ";
double user_input;
cin >> user_input;
int user_input_to_int = int(user_input);
double decimal_value = user_input - user_input_to_int;
ostringstream oss;
oss << std::noshowpoint << decimal_value;
string num_str = oss.str();
int str_length = num_str.size()-2;
int multiplier = 1;
for (int x = 0; x < str_length; x++)
{
multiplier *= 10;
}
cout << "\n";
cout << "Whole number: " << user_input_to_int << endl;
cout << "Decimal value: " << decimal_value*multiplier << endl;
}
Compare the difference between double and integer part. It is working only if x is less than 2^63.
while (x - long long(x) > 0)
find 2 real numbers find the fractional part smaller than these numbers
I'm writing a C++ program that converts a decimal number to binary and hexadecimal.
The problem is that for some reason it concatenates the number "1875954912" to both representations every time.
I've tried a bunch of things - mainly changing up how the program calculates numArrayLength and the for-loop in my decToBase function, but I haven't been able to figure out why this happens yet.
The program is not complete by the way - it doesn't turn integers bigger than 9 into letters for the hex representation yet, but that's not my main concern right now.
Here is my code:
#include <iostream>
#include <cmath>
using namespace std;
int howManyBitsNeeded(int someNum, int base) {
int numOfDivisions = 0;
while (someNum != 0) {
someNum = floor(someNum / base);
numOfDivisions += 1;
}
return numOfDivisions;
}
int decToBase(int someNum, int base) {
int bitsNeeded = howManyBitsNeeded(someNum,base);
int numArrayLength = bitsNeeded;
int numArray[bitsNeeded];
while (bitsNeeded > 0) {
numArray[bitsNeeded] = (someNum % base);
someNum = floor(someNum / base);
bitsNeeded -= 1;
}
for (int k = (numArrayLength-1); k >= 0; --k) {
cout << numArray[(numArrayLength - k)];
}
}
int main() {
int inpNum;
cout << "Enter your number: ";
cin >> inpNum;
cout << "Binary representation: " << decToBase(inpNum,2) << endl;
cout << "Hexadecimal representation: " << decToBase(inpNum,16);
return 0;
}
And here's what the output looks like:
Enter your number: 25
Binary representation: 110011875954912
Hexadecimal representation: 191875954912
Any help would be greatly appreciated!
Your decToBase is declared as returning an int, but it doesn't actually return anything. Your compiler should warn you about this. Since you're not returning anything here, change its return type to void. Then instead of trying to print its return value, simply call the function without printing it:
std::cout << "Binary representation: ";
decToBase(inpNum, 2); // this already prints the number, no need to pass it to std::cout
std::cout << endl;
std::cout << "Hexadecimal representation: ";
decToBase(inpNum, 16);
std::cout << std::endl;
Or of course you can change the function to return the string that you want to print instead of printing it inside the function.
Also, there's an issue here:
int numArray[bitsNeeded];
This is out of range when you try to access it here:
while (bitsNeeded > 0) {
numArray[bitsNeeded] = (someNum % base);
And also later when you try to print it. To get rid of this off by one error, you have to change this to
numArray[bitsNeeded-1] = (someNum % base);
And in the output change it to
cout << numArray[(numArrayLength - k -1)];
And while you're at it, instead of having it as a VLA (which isn't part of C++ and only works if the compiler tolerates it), I would recommend a vector:
std::vector<int> numArray(bitsNeeded+1); // include <vector> for this to work
Furthermore, note that integer division is already truncated, so unless you plan to support negative numbers later on, you can silence a warning about implicit double to int conversion by changing this:
someNum = floor(someNum / base);
To this:
someNum /= base;
So, I hate to ask, but, I'm having some issue with this, I'm new to C++ and I'm just starting out. Everything is done for the most part. Expect for a little thing.
Line 35-36 should be calculating the average (Which for some reason, I haven't been able to get it to work.)
Line 41-47 should print out the percentage that heads/tails was landed on with precision to one decimal, and then print out the correct numbers of * to represent the percentage.
But, when I run it, my heads/tail count is messed up. As well as my percentage numbers. I'm just looking for a push in the right direction.
#include <cstdlib>
#include <iostream>
#include <ctime>
#include <iomanip>
using std::cout; using std::cin; using std::endl;
using std::fixed; using std::setprecision;
int main()
{
srand(time(0));
int userInput,
toss,
headsCount,
tailsCount;
double headsPercent = 0,
tailsPercent = 0;
cout << "How many times do you want to toss the coin? ";
cin >> userInput;
while(userInput < 0)
{
cout << "Please enter a positive number: ";
cin >> userInput;
}
for(int i = 1; i < userInput; i++)
{
toss = rand() % 2;
if(toss == 0)
headsCount++;
else
tailsCount++;
}
headsPercent = userInput / headsCount * 100;
tailsPercent = userInput / tailsCount;
cout << "Heads: " << headsCount << endl
<< "Tails: " << tailsCount << endl << endl;
cout << "Heads Percentage: " << fixed << setprecision(1) << headsPercent << " ";
for(int b = 0; b < headsPercent; b++)
cout << "*";
cout << "\nTails Percentage: " << tailsPercent << " ";
for(int b = 0; b < tailsPercent; b++)
cout << "*";
return 0;
}
In addition to the uninitialized variables here, that others have pointed out, the calculations are all wrong.
Take out paper and pencil, and run some your own calculations the old-fashioned way.
Let's say there were five tosses, three heads, two tails. This means that (after fixing the uninitialized variables):
userInput=5
headsCount=3
tailsCount=2
Now, here's how you're calculating your supposed percentages:
headsPercent = userInput / headsCount * 100;
tailsPercent = userInput / tailsCount;
So, using your own numbers, you will get:
headsPercent = 5 / 3 * 100
tailsPercent = 5 / 2;
Does this look right to you? Of course not. You can do the arithmetic yourself. Divide 5 by 3 and multiply by 100. This is integer division, so five divided by three is 1, multiplied by 100 is 100. Five divided by two is two. So you get 100% and 2% here.
Of course, that's wrong. Two and three times, out of five, is 40% and 60%, respectively.
Writing a program means:
A) Figure out how calculations need to be made
B) Write the code to do the calculations.
You're still on step A. You need to figure out how you want to make these calculations so they're correct, first.
This has nothing really to do with C++. If you were using any other language, and coded this, in that manner, you'll get the same wrong answers.
The only thing this might have to do with C++ is that integer division, in C++ does not produce a fractional amount. It's integer division. But that's not your only problem.
Firstly u have to correct ur basics of mathematics.
Calculating %age means
example
(Marks obtained)/(Total marks)*100
Not (Total marks/marks obt)*100
Dividing any no by 0 is not defined. So if ur current code randomly assign toss or head =0, then obviously u will have errors.
Secondly talking about codes, U should either initialize i from 0 , or u should use
for (i=1; i<=userInput; i++)
As otherwise the head+toss value will be userInput-1.
Also remember to initialise variables like
Int headsCount=0;
etc. As the variable will take any random value if not initialised to a fixed no. (Though it does not creates a problem here)
And just change the datatype
int userInput,
toss,
headsCount,
tailsCount;
To
double userInput,
toss,
headsCount,
tailsCount;
This will solve your problem.
Advice: Please use
using namespace std;
in the starting of ur programs as u have to type a lot of std::
Welcome to C++. You need to initialise your variables. Your compiler should have warned you that you were using a variable without initialising it. When you don't initialise a value, your program has undefined behaviour.
I'm talking about headsCount and tailsCount. Something like this should be fine:
int headsCount = 0, tailsCount = 0;
Also note that your loop should start at 0, not 1, since you are using the < operator on the final condition.
Finally, your percentage calculations are backwards. It should be:
headsPercent = headsCount * 100 / userInput;
tailsPercent = tailsCount * 100 / userInput;
Now, there's a weird thing that might happen because you are using integer division. That is, your percentages might not add up to 100. What's happening here is integer truncation. Note that I dealt with some of this implicitly using the 100x scale first.
Or, since the percentages themselves are double, you can force the calculation to be double by casting one of the operands, thus avoiding integer truncation:
headsPercent = static_cast<double>(headsCount) / userInput * 100;
In fact, since the only two possibilities are heads and tails, you only need to count one of them. Then you can do:
tailsPercent = 100 - headsPercent;
1) This loop should start from 0:
for(int i = 1; i < userInput; i++)
2) The divisions are not correct:
//headsPercent = userInput / headsCount * 100;
//tailsPercent = userInput / tailsCount;
headsPercent = headsCount / userInput * 100;
tailsPercent = tailsCount / userInput * 100;
3) Finally:
cout << "\nTails Percentage: " << fixed << setprecision(1) << tailsPercent << " ";
Whilst working on a personal project of mine, I came across a need to divide two very large arbitrary numbers (each number having roughly 100 digits).
So i wrote out the very basic code for division (i.e., answer = a/b, where a and b are imputed by the user)and quickly discovered that it only has a precision of 16 digits! It may be obvious at this point that Im not a coder!
So i searched the internet and found a code that, as far as i can tell, uses the traditional method of long division by making a string(but too be honest im not sure as im quite confused by it). But upon running the code it gives out some incorrect answers and wont work at all if a>b.
Im not even sure if there's a better way to solve this problem than the method in the code below!? Maybe there's a simpler code??
So basically i need help to write a code, in C++, to divide two very large numbers.
Any help or suggestions are greatly appreciated!
#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std; //avoids having to use std:: with cout/cin
int main (int argc, char **argv)
{
string dividend, divisor, difference, a, b, s, tempstring = ""; // a and b used to store dividend and divisor.
int quotient, inta, intb, diff, tempint = 0;
char d;
quotient = 0;
cout << "Enter the dividend? "; //larger number (on top)
cin >> a;
cout << "Enter the divisor? "; //smaller number (on bottom)
cin >> b;
//making the strings the same length by adding 0's to the beggining of string.
while (a.length() < b.length()) a = '0'+a; // a has less digits than b add 0's
while (b.length() < a.length()) b = '0'+b; // b has less digits than a add 0's
inta = a[0]-'0'; // getting first digit in both strings
intb = b[0]-'0';
//if a<b print remainder out (a) and return 0
if (inta < intb)
{
cout << "Quotient: 0 " << endl << "Remainder: " << a << endl;
}
else
{
a = '0'+a;
b = '0'+b;
diff = intb;
//s = b;
// while ( s >= b )
do
{
for (int i = a.length()-1; i>=0; i--) // do subtraction until end of string
{
inta = a[i]-'0'; // converting ascii to int, used for munipulation
intb = b[i]-'0';
if (inta < intb) // borrow if needed
{
a[i-1]--; //borrow from next digit
a[i] += 10;
}
diff = a[i] - b[i];
char d = diff+'0';
s = d + s; //this + is appending two strings, not performing addition.
}
quotient++;
a = s;
// strcpy (a, s);
}
while (s >= b); // fails after dividing 3 x's
cout << "s string: " << s << endl;
cout << "a string: " << a << endl;
cout << "Quotient: " << quotient << endl;
//cout << "Remainder: " << s << endl;
}
system ("pause");
return 0;
cin.get(); // allows the user to enter variable without instantly ending the program
cin.get(); // allows the user to enter variable without instantly ending the program
}
There are much better methods than that. This subtractive method is arbitrarily slow for large dividends and small divisors. The canonical method is given as Algorithm D in Knuth, D.E., The Art of Computer Programming, volume 2, but I'm sure you will find it online. I'd be astonished if it wasn't in Wikipedia somewhere.
This small program is made to figure out the first and second digits of a 2 digit number. However, when I try using it on the number 99 then it prints 9 and 8, but other 2 digit numbers seem to work fine though. This is probably trivial, but I'm relatively new at programming.
#include <iostream>
using namespace std;
void test(int num) {
float numValue = (num*1.0) / 10;
cout << numValue << endl;
// prints 9.9
int firstDigit = num / 10;
cout << firstDigit << endl;
// prints 9
int secondDigit = (numValue - firstDigit) * 10;
cout << secondDigit << endl;
// prints 8, supposed to be (9.9 - 9) * 10
}
int main()
{
test(99);
return 0;
}
That happens because (numValue - firstDigit) is not exactly 0.9, but rather 0.89999..., because floating-point numbers are not accurate generally. So when you multiply 0.8999... by 10, you get the result of 8.999... However, then you are putting it into an int variable, so it gets trimmed and becomes exactly 8.
You don't really need floating point arithmetic for your exact task though, using integers will be enough.