Handling large numbers in C++? - c++

What is the best way to handle large numeric inputs in C++ (for example 10^100)?
For algorithms I usually switch over to ruby and I sometimes use strings.
Any other good methods?

It sounds like you're looking for a way to enter Arbitrary Precision numbers.
here are two libraries you could use: GMP and MAPM

Check out The Large Integer Case Study in C++.pdf by Owen Astrachan. I found this file extremely useful with detail introduction and code implementation. It doesn't use any 3rd-party library. I have used this to handle huge numbers (as long as you have enough memory to store vector<char>) with no problems.
Idea:
It implements an arbitrary precision integer class by storing big int in a vector<char>.
vector<char> myDigits; // stores all digits of number
Then all operations related to the big int, including <<, >>, +, -, *, ==, <, !=, >, etc., can be done based on operations on this char array.
Taste of the code:
Here is the header file, you can find its cpp with codes in the pdf file.
#include <iostream>
#include <string> // for strings
#include <vector> // for sequence of digits
using namespace std;
class BigInt
{
public:
BigInt(); // default constructor, value = 0
BigInt(int); // assign an integer value
BigInt(const string &); // assign a string
// may need these in alternative implementation
// BigInt(const BigInt &); // copy constructor
// ~BigInt(); // destructor
// const BigInt & operator = (const BigInt &);
// assignment operator
// operators: arithmetic, relational
const BigInt & operator += (const BigInt &);
const BigInt & operator -= (const BigInt &);
const BigInt & operator *= (const BigInt &);
const BigInt & operator *= (int num);
string ToString() const; // convert to string
int ToInt() const; // convert to int
double ToDouble() const; // convert to double
// facilitate operators ==, <, << without friends
bool Equal(const BigInt & rhs) const;
bool LessThan(const BigInt & rhs) const;
void Print(ostream & os) const;
private:
// other helper functions
bool IsNegative() const; // return true iff number is negative
bool IsPositive() const; // return true iff number is positive
int NumDigits() const; // return # digits in number
int GetDigit(int k) const;
void AddSigDigit(int value);
void ChangeDigit(int k, int value);
void Normalize();
// private state/instance variables
enum Sign{positive,negative};
Sign mySign; // is number positive or negative
vector<char> myDigits; // stores all digits of number
int myNumDigits; // stores # of digits of number
};
// free functions
ostream & operator <<(ostream &, const BigInt &);
istream & operator >>(istream &, BigInt &);
BigInt operator +(const BigInt & lhs, const BigInt & rhs);
BigInt operator -(const BigInt & lhs, const BigInt & rhs);
BigInt operator *(const BigInt & lhs, const BigInt & rhs);
BigInt operator *(const BigInt & lhs, int num);
BigInt operator *(int num, const BigInt & rhs);
bool operator == (const BigInt & lhs, const BigInt & rhs);
bool operator < (const BigInt & lhs, const BigInt & rhs);
bool operator != (const BigInt & lhs, const BigInt & rhs);
bool operator > (const BigInt & lhs, const BigInt & rhs);
bool operator >= (const BigInt & lhs, const BigInt & rhs);
bool operator <= (const BigInt & lhs, const BigInt & rhs);

If you wish to make your own code for the purpose try using strings to store big numbers... you can then create basic ops like + - / * on them... for example -
#include <iostream>
using namespace std;
string add (string &s1, string &s2){
int carry=0,sum,i;
string min=s1,
max=s2,
result = "";
if (s1.length()>s2.length()){
max = s1;
min = s2;
} else {
max = s2;
min = s1;
}
for (i = min.length()-1; i>=0; i--){
sum = min[i] + max[i + max.length() - min.length()] + carry - 2*'0';
carry = sum/10;
sum %=10;
result = (char)(sum + '0') + result;
}
i = max.length() - min.length()-1;
while (i>=0){
sum = max[i] + carry - '0';
carry = sum/10;
sum%=10;
result = (char)(sum + '0') + result;
i--;
}
if (carry!=0){
result = (char)(carry + '0') + result;
}
return result;
}
int main (){
string a,b;
cin >> a >> b;
cout << add (a,b)<<endl;
return 0;
}

Are you looking for how to perform operations on the large inputs you receive? There is a big integer C++ library (similar to Java) that allows you to perform arithmetic operations...

assuming you are talking about inputting numbers, double precision would get you up to 1.7976931348623157 x 10^308

You might want to have a look to gmplib, an arbitrary precision number handling library for C and C++

If you want it to be accurate, you need a library made to deal with big numbers. Java has BigInt that will always be accurate no matter how many digits you want to take it to, and provides math operations on them. All the source code is included, you could transfer it, but this really isn't the kind of thing C++ is best at--I'd use a JVM based language and use one of the Big libraries.
I don't think I'd use ruby for this unless you wanted it to be slow, and I'm assuming that since you are talking about C++, speed is somewhat of a design consideration.

As others have already pointed out, there are various bignum/arbitrary precision libraries in C++ that you would likely find useful. If speed isn't necessary, I'm under the impression that Python and Lisp both use bignums by default.

Consider boost::cpp_int
#include <boost/multiprecision/cpp_int.hpp>
#include <iostream>
int main()
{
using namespace boost::multiprecision;
cpp_int u = 1;
for(unsigned i = 1; i <= 100; ++i)
u *= i;
// prints 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000 (i.e. 100!)
std::cout << u << std::endl;
return 0;
}

Well I think the best way to do such arithmetic calculation is by using strings. Give input as command line arguments and then manipulate the whole logic using string functions like atoi() and itoa()! But, hey can this be done for multiplication and Division? I think in this way strlen of strings entered doesn't matter for programming for compiler until the logic is fine.

Related

c++ operator overload with string comparing

hello i have problem in my school c++ lab, my bool operator > should be return true if lhs is greater than rhs, however it always return false. i try print out lhs.tostring(), it show the number correctly.
my lhs and rhs is a string value.
due to some confidence restrict from my school work, i am not allow to post all the function of my work.
Updated information: ithis lab only can use c++14 and can't include any additional lib. The int value is written in string, and need to compare which is bigger. Assuming there is no negative and any letter other than number
some part of my header file
#include <cstring>
#include <iostream>
namespace CS170
{
class BigNum
{
public:
/* Constructor of BigNum object.
Takes in a character string and
constructs a BigNum */
BigNum(const char * rhs = "0");
/* one of rule of 3 need destructor */
~BigNum();
/* Return a character pointer pointing
to the start of the array representing the big num */
const char * toString() const;
/* Return how many digits the number has */
size_t getNumDigits() const;
BigNum & operator =(const BigNum & rhs);
private:
size_t len;
char* num;
};
}
bool operator >(const CS170::BigNum &lhs, const CS170::BigNum &rhs);
cpp
namespace CS170
{
BigNum::BigNum(const char * rhs )
:len{strlen(rhs)}, num{new char[len+1]}
{
strcpy(num,rhs);
}
BigNum::~BigNum()
{
}
const char * BigNum::toString() const
{
return num;
}
size_t BigNum::getNumDigits() const
{
return len;
}
}
bool operator >(const CS170::BigNum &lhs, const CS170::BigNum &rhs)
{
CS170::BigNum left_value{lhs};
CS170::BigNum right_value{rhs};
std::cout << std::endl;
std::cout << left_value.toString() << " " << right_value.toString() <<
std::endl;
/*this don't work for comparing**/
if(left_value.toString() > right_value.toString())
return true;
else
return false;
}
left_value.toString() > right_value.toString()
This does not do what you think it does. toString() returns a const char*, a pointer to some data. Formally the behaviour of > in your case is undefined since the pointers are not part of the same array, and even if they were, the result would not depend on the string contents.
To check the lexicogrammatical order of strings, you should use the right tool for it, for instance std::string::operator>:
std::string lhs_string{left_value.toString()};
std::string rhs_string{rght_value.toString()};
if (lhs_string > rhs_string)
// ...
// note: here you could simply do return lhs_string > rhs_string;
If you're using a recent compiler and C++17 is an option, you could also use those tools without copying data around:
#include <string_view>
const char* lhs = "programming";
const char* rhs = "language";
std::string_view lhs_string{lhs};
std::string_view rhs_string{rhs};
lhs_string>rhs_string // lexicogrammatical order
live demo
const char* cannot be compared in the way you are trying to. You have to use strcmp. Example usage would look like:
if (strcmp(left_value.toString(), right_value.toString()) > 0)
{
return true;
}
The last part of the function could even be simplified to:
return strcmp(left_value.toString(), right_value.toString()) > 0;
is nearly there but i don't work if compare 11 > 2, as is still read only the first string.
bool operator >(const CS170::BigNum &lhs, const CS170::BigNum &rhs)
{
CS170::BigNum left_data{lhs};
CS170::BigNum right_data{rhs};
int result = strncmp(left_data.toString(), right_data.toString(),20);
return result > 0;
}

how to convert a built-in type to user-defined type

I have a class called BigInteger which supports big integer operation. I want to implement mixture operation between BigInteger and built-in type 'int'. In other word, I want to support following statements
BigInteger a(10);
a + 10;
10 + a;
I know overloaded function can deal with it
BigInteger operator +(const BigInteger&, const int&);
BigInteger operator +(const int&, const BigInteger&);
Besides, I know a conversion operator can only deal with it ,
operator int();
But the above function support convert BigInteger to int which will lose precision. I am looking for some methods which will be simpler than overloaded function and keep precision.
Thanks, everyone.
I try it,
#include <iostream>
using namespace std;
class BigInteger
{
public:
BigInteger(const int& i)
{
cout << "construct: " << i << endl;
val = i;
}
// BigInteger operator +(const BigInteger& tmp) const
// {
// return BigInteger(val + tmp.val);
// }
friend ostream& operator <<(ostream& os, const BigInteger& bi)
{
os << bi.val << endl;
return os;
}
int val;
};
BigInteger operator +(const BigInteger& a, const BigInteger& b)
{
return BigInteger(a.val + b.val);
}
int main(int argc, const char *argv[])
{
BigInteger a(12);
cout << (a + 123) << endl;
cout << (1231 + a) << endl;
return 0;
}
why can't I use member function? How it works?
You need to add constructor that will take value of BigInteger from int
BigInteger (const int& value)
So, make a constructor BigInteger(int), and define an BigInteger operator(const BigInteger &lhs, const BigInteger &rhs).
You should implement overloads for BigInteger operator+(BigInteger const&, int); As Herb Sutter and Andrei Alexandrescu wrote in C++ Coding Standarts (Item 29) you should "Consider overloading to avoid implicit type conversation". Your BigInteger might use operator new for construction wich can be avoided in this particular case. Additionally you should implement the binary operation in terms of unary operation:
class BigInteger {
// ..
public:
BigInteger& operator+=(int); // this operation can reuse BigIntegers buffer
// ..
}
BigInteger operator+(BigInteger lhs, int rhs)
{
return lhs+=rhs;
}
BigInteger operator+(int lhs, BigInteger rhs)
{
return rhs+=lhs;
}
Make sure that you are only ever converting from int to BigInteger, never the reverse. Or, throw an exception if the reverse conversion would lead to an overflow.
define the constructor BigInteger(int) and overload the operator BigInteger operator+(const BigInteger &left, const BigInteger &right).

Overloading Operator + in C++

Ok, I am working through a book and trying to learn C++ operator overloading. I created a BigInt class that takes a single int (initially set to 0) for the constructor. I overloaded the += method and it works just fine in the following code:
BigInt x = BigInt(2);
x += x;
x.print( cout );
The code will output 4. So, then I was working on overloading the global operator + using the following code:
BigInt operator+(const BigInt lhs, const BigInt rhs)
{
BigInt returnValue(lhs);
returnValue += rhs;
return returnValue;
}
This also works fine for the following code:
BigInt x = BigInt(1);
BigInt y = BigInt(5);
BigInt z = x + y;
z.print();
This prints out 6. However, when I try to execute the following code, it just doesn't work. The book doesn't explain very well and implies that it should simply work.
BigInt x = BigInt(1);
BigInt z = x + 5;
z.print();
This prints out 1. I'm not sure why z is 1 when it should be 6. I googled online and on stackoverflow but I couldn't find anyone else that was having a problem exactly like this. some were close, but the answers just didn't fit. Any help is much appreciated!
most likely problem is in += operator. Post code for it.
You need an overload for adding an int to BigInt; the constant 5 in your example is of type int, not BigInt. Something like this should work:
BigInt operator+(const BigInt lhs, const int rhs)
{
BigInt returnValue(rhs);
returnValue += lhs;
return returnValue;
}
You might want one for operator+(const int lhs, const BigInt rhs) too.
The following super-simplified code (the minimum I can add to include all your code and make it into a valid stand-alone executable program):
#include <iostream>
class BigInt
{
public:
BigInt(int i): _i(i) {}
void print() { std::cout << "BigInt(" << _i << ")\n"; }
void operator +=(const BigInt rhs) { _i += rhs._i; }
private:
int _i;
};
BigInt operator+(const BigInt lhs, const BigInt rhs)
{
BigInt returnValue(lhs);
returnValue += rhs;
return returnValue;
}
int main() {
BigInt x = BigInt(1);
BigInt y = BigInt(5);
BigInt z = x + y;
z.print();
BigInt ax = BigInt(1);
BigInt az = ax + 5;
az.print();
return 0;
}
emits, as predictable:
BigInt(6)
BigInt(6)
Please make the minimum possible alterations to this working code to reproduce the bug you observe -- that will of course show where your bug exactly lies.
The code you've posted looks fine and should work. Problems you're seeing are almost certainly due to the copy constructor or assignment operator of your BigInt class.

error C2593: 'operator +' is ambiguous

If I have the following files, I get this error (c2593 in VC9).
If I un-comment the prototype in main.cpp, the error disappears. I need to maintain the same functionality while keeping the class out of main.cpp. How can I do that?
Thanks.
main.cpp:
#include "number.h"
//const Number operator + (const Number & lhs, const Number & rhs);
int main(void)
{
Number n1(2); // n1 = 2
Number n2(9,3); // n2 = 3
Number n3 = n1+n2; // n3 = 5
}
number.h:
struct Number
{
int num;
Number(int n=0,int d=1) {num = n/d;}
operator int() {return num;}
operator double() {return num*1.0;}
};
number.cpp:
#include "Number.h"
const Number operator + (const Number & lhs, const Number & rhs)
{
Number tmp;
tmp.num = lhs.num + rhs.num;
return tmp;
}
Try putting the prototype in the Number header file:
number.h:
struct Number
{
int num;
Number(int n=0,int d=1) {num = n/d;}
operator int() {return num;}
operator double() {return num*1.0;}
};
const Number operator + (const Number & lhs, const Number & rhs);
number.cpp:
#include "Number.h"
const Number operator + (const Number & lhs, const Number & rhs)
{
Number tmp;
tmp.num = lhs.num + rhs.num;
return tmp;
}
main.cpp:
#include "number.h"
int main(void)
{
Number n1(2); // n1 = 2
Number n2(9,3); // n2 = 3
Number n3 = n1+n2; // n3 = 5
}
You never declare operator + in number.h, you only define it in number.cpp - therefore, when you include number.h in main.cpp, it doesn't know where to go to find operator +.
You must put the declaration of operator + in number.h, outside of the class, then define it in number.cpp
That commented line should go in number.h
EDIT: in number.h but as a free function.
One other thing to note with your code before you go back to us with a very similar question: Better remove the operator int and operator double functions. They will cause you major headache. Let's make a few examples:
Number a, b;
1 + b;
// ambiguous: operator+(int, int) or
// operator+(Number, Number) ?
// did you intend to use those for this case?
float x = a;
// ambiguous: from int -> float or
// double -> float ?
In the original situation you had, your addition was ambiguous, because there were operator+(double, double) and operator+(int, int) builtin operators considered and they were equally well. Others solved that problem. But before you start and run into these other problems, better remove the conversion functions and insert explicit functions like asDouble or something similar.
Apart from other's answers to declare operator+ in the header file, I suggest you to have operator+= in your struct as well.
struct Number
{
// your other declarations.
Number& operator+=(const Number& other)
{
this->num += other.num;
return *this;
}
};
const Number operator+(const Number& lhs, const Number& rhs)
{
Number ret(lhs);
ret += rhs;
return ret;
}
This way it is efficient to call x += y; instead of x = x + y;
Personally I like it better to declare the operators inside the class:
struct Number
{
int num;
Number(int n=0,int d=1) {num = n/d;}
operator int() {return num;}
operator double() {return num*1.0;}
Number operator+(const Number &arg) const;
};
and then:
Number Number::operator+(const Number &arg)
{
...
}

Way of overloading operator without changing original values?

I'm wondering if you can overload an operator and use it without changing the object's original values.
Edited code example:
class Rational{
public:
Rational(double n, double d):numerator_(n), denominator_(d){};
Rational(){}; // default constructor
double numerator() const { return numerator_; } // accessor
double denominator() const { return denominator_; } // accessor
private:
double numerator_;
double denominator_;
};
const Rational operator+(const Rational& a, const Rational& b)
{
Rational tmp;
tmp.denominator_ = (a.denominator() * b.denominator());
tmp.numerator_ = (a.numerator() * b.denominator());
tmp.numerator_ += (b.numerator() * a.denominator());
return tmp;
}
I made the accessors const methods, but I'm still getting a privacy error for every tmp.denominator_ / numerator_.
Maybe I'm missing something, but why don't you just take out the code that modifies the arguments?
const Rational Rational::operator+(Rational& num)
{
Rational tmp;
tmp.denominator_ = (denominator_*num.denominator_);
//numerator_*=num.denominator_;
//num.numerator_*=denominator_;
tmp.numerator_ = (numerator_+num.numerator_);
return tmp;
}
This would be caught earlier by being const-correct.
That means your function signature should be this:
Rational Rational::operator+(const Rational& num) const
Then you will get errors because you are modifying const objects. The way your operators are written now is generally considered incorrect.
When you add 2 + 3, neither 2 nor 3 changes: they are const.
Edit
Sorry, I missed the actual math part. Here are a few things:
As a member function (what I have above), do this:
// Note that I use lhs and rhs to refer to the left-hand
// and right-hand sides of an operation. As a member function
// my left-hand side is implicitly `this`.
Rational Rational::operator+(const Rational& rhs) const
{
Rational temp;
temp.denominator_ = (denominator() * rhs.denominator());
temp.numerator_ = (numerator() * rhs.denominator());
temp.numerator_ += (denominator() * rhs.numerator());
return temp;
}
As a global function, do this:
Rational operator+(const Rational& lhs, const Rational& rhs)
{
Rational temp;
temp.denominator_ = (lhs.denominator() * rhs.denominator());
temp.numerator_ = (lhs.numerator() * rhs.denominator());
temp.numerator_ += (lhs.denominator() * rhs.numerator());
return temp;
}
The issue here is you'll get access violations: your member variables are private to the class. You need to let the compiler know that it is okay if this function handles your class's private variables by making the function a friend of the class:
class Rational
{
public:
friend Rational operator+(const Rational& lhs, const Rational& rhs);
}
What you're looking for are the "binary" addition and subtraction operators:
const Rational operator+(const Rational& A, const Rational& B)
{
Rational result;
...
return result;
}
update (in response to new code and comments):
You are getting that error because your accessor functions are not declared as constant functions, so the compiler has to assume that they might modify the original object. Change your accessors as follows, and you should be good to go:
double numerator() const { return numerator_; }
double denominator() const { return denominator_; }
update
To properly handle privacy issues, you should declare the binary operator+ function as a friend of the Rational class. Here is how it would look:
class Rational {
public:
Rational(double n, double d):numerator_(n), denominator_(d) {};
Rational() {}; // default constructor
double numerator() const { return numerator_; } // accessor
double denominator() const { return denominator_; } // accessor
friend Rational operator+(const Rational& A, const Rational& B);
private:
double numerator_;
double denominator_;
};
const Rational operator+(const Rational& a, const Rational& b)
{
Rational result;
result.denominator_ = (a.denominator_ * b.denominator_);
result.numerator_ = (a.numerator_ * b.denominator_);
result.numerator_ += (b.numerator_ * a.denominator_);
return result;
}
no .. you have to rewrite + and - .. it's not that hard. Then change the signatures to
const Rational Rational::operator+(Rational& num) const
const Rational Rational::operator-(Rational& num) const
that way the compiler will let you know if you are modifying your object.
Since your class already provides accessors to the numerator and denominator, and it has a public constructor, there is no need to use any friends. You can overload operator+ as follows:
const Rational operator+(const Rational& rhs, const Rational& lhs)
{
double newNumerator = rhs.numerator() * lhs.denominator() +
rhs.denominator() * lhs.numerator();
return Rational(newNumerator, rhs.denominator() * lhs.denominator());
}
In response to some of the other answers:
Everything about this question is exactly answered by Item 24 of Effective C++ (Third Edition). In this item, Scott Meyers shows that, in cases dealing with numerical types where you want to support implicit conversion in an intuitive manner, it is best to use a non-member non-friend function.
Say your Rational class had the following constructor:
Rational::Rational(double numerator = 0, double denominator = 1);
In this case, if operator+ were a member function, trying to do mixed mode arithmetic would work only half the time:
Rational oneHalf(1, 2);
oneHalf + 2; // works
2 + oneHalf; // error!
In the second example, operator+ for integers is called. The way to fix this is to make operator+ for Rationals a non-member function as shown above.
If you have access to Effective C++, you should also check out Item 23: Prefer non-member non-friend functions to member functions.
I prefer to implement operator+ in terms of operator+=. Then clients get to choose if they want a temp copy or not. Also operator+= is more natural as a member function so it has access to the private parts.
class Rational {
public:
Rational(double n, double d):numerator_(n), denominator_(d) {};
Rational() {}; // default constructor
double numerator() const { return numerator_; } // accessor
double denominator() const { return denominator_; } // accessor
Rational & operator+=(const Rational& b) {
denominator_ *= b.denominator_);
numerator_ *= (b.denominator_);
numerator_ += (b.numerator_ * denominator_);
return *this;
}
private:
double numerator_;
double denominator_;
};
const Rational operator+(const Rational& a, const Rational& b)
{
Rational result(a);
result += b;
return result;
}
However the easiest way to implement operator overloading is to use the boost operators library. Then you implement the minimum set of operators and boost take care of the rest and all the gotchas (including wintermute's example of 2 + oneHalf.