I am trying to better understand how 'big numbers' libraries work, (like GMP for example).
I want to write my own function to Add() / Subtract() / Multiply() / Divide()
The class is traditionally defined ...
std::vector<unsigned char> _numbers; // all the numbers
bool _neg; // positive or negative number
long _decimalPos; // where the decimal point is located
// so 10.5 would be 1
// 10.25 would be 2
// 10 would be 0 for example
First I need to normalise the numbers so I can do
Using 2 numbers
10(x) + 10.25(y) = 20.25
For simplicity, I would make them the same length,
For x:
_numbers = (1,0,0,0) decimal = 2
For y:
_numbers = (1,0,2,5) decimal = 2
And I can then reverse add x to y in a loop
...
// where x is 10.00 and y is 10.25
...
unsigned char carryOver = 0;
int totalLen = x._numbers.size();
for (size_t i = totalLen; i > 1 ; --i )
{
unsigned char sum = x._numbers[i-1] + y._numbers[i-1] + carryOver;
carryOver = 0;
if (sum > _base)
{
sum -= _base;
carryOver = 1;
}
numbers.insert( number.begin(), sum);
}
// any left over?
if (carryOver > 0)
{
numbers.insert( number.begin(), 1 );
}
// decimal pos is the same for this number as x and y
...
The example above will work for adding two positive numbers, but will soon fall over once I need to add a negative number to a positive number.
And this gets more complicated when it comes to subtracting numbers, then even worse for multiplications and divisions.
Can someone suggest some simple functions to Add() / Subtract() / Multiply() / Divide()
I am not trying to re-write / improve libraries, I just want to understand how they work with numbers.
addition and substractions are pretty straightforward
You need to inspect signs and magnitudes of operands and if needed convert the operation to/from +/-. Typical C++ implementation of mine for this is like this:
//---------------------------------------------------------------------------
arbnum arbnum::operator + (const arbnum &x)
{
arbnum c;
// you can skip this if you do not have NaN or Inf support
// this just handles cases like adding inf or NaN or zero
if ( isnan() ) return *this;
if (x.isnan() ) { c.nan(); return c; }
if ( iszero()) { c=x; return c; }
if (x.iszero()) return *this;
if ( isinf() ) { if (x.isinf()) { if (sig==x.sig) return *this;
c.nan(); return c; } return *this; }
if (x.isinf()) { c.inf(); return c; }
// this compares the sign bits if both signs are the same it is addition
if (sig*x.sig>0) { c.add(x,this[0]); c.sig=sig; }
// if not
else{
// compare absolute values (magnitudes)
if (c.geq(this[0],x)) // |this| >= |x| ... return (this-x)
{
c.sub(this[0],x);
c.sig=sig; // use sign of the abs greater operand
}
else { // else return (x-this)
c.sub(x,this[0]);
c.sig=x.sig;
}
}
return c;
}
//---------------------------------------------------------------------------
arbnum arbnum::operator - (const arbnum &x)
{
arbnum c;
if ( isnan() ) return *this;
if (x.isnan() ) { c.nan(); return c; }
if ( iszero()) { c=x; c.sig=-x.sig; return c; }
if (x.iszero()) return *this;
if ( isinf() ) { if (x.isinf()) { if (sig!=x.sig) return *this;
c.nan(); return c; } return *this; }
if (x.isinf()) { c.inf(); c.sig=-x.sig; return c; }
if (x.sig*sig<0) { c.add(x,this[0]); c.sig=sig; }
else{
if (c.geq(this[0],x))
{
c.sub(this[0],x);
c.sig=sig;
}
else {
c.sub(x,this[0]);
c.sig=-x.sig;
}
}
return c;
}
//---------------------------------------------------------------------------
where:
geq is unsigned comparison greater or equal
add is unsigned +
sub is unsigned -
division is a bit more complicated
see:
bignum divisions
approximational bignum divider
For divisions you need to have already implemented things like +,-,*,<<,>> and for some more advanced approaches you need even things like: absolute comparison (you need them for +/- anyway) , sqr, number of used bits usually separate for fractional and integer part.
The most important is the multiplication see Fast bignum square computation because it is core for most division algorithms.
performance
for some hints see BigInteger numbers implementation and performance
text conversion
If your number is in ASCII or in BASE=10^n digits then this is easy but If you use BASE=2^n instead for performance reasons then you need to have fast functions capable of converting between dec and hex strings so you can actually load and print some numbers to/from your class. see:
How do I convert a very long binary number to decimal?
How to convert a gi-normous integer (in string format) to hex format?
Related
My problem seems to be pretty simple: I wrote a program that manually adds floating point numbers together. This program has certain restrictions. (such as no iostream or use of any unary operators), so that is the reason for the lack of those things. As for the problem, the program seems to function correctly when adding two positive floats (1.5 + 1.5 = 3.0, for example), but when adding two negative numbers (10.0 + -5.0) I get very wacky numbers. Here is the code:
#include <cstdio>
#define BIAS32 127
struct Real
{
//sign bit
int sign;
//UNBIASED exponent
long exponent;
//Fraction including implied 1. at bit index 23
unsigned long fraction;
};
Real Decode(int float_value);
int Encode(Real real_value);
Real Normalize(Real value);
Real Add(Real left, Real right);
unsigned long Add(unsigned long leftop, unsigned long rightop);
unsigned long Multiply(unsigned long leftop, unsigned long rightop);
void alignExponents(Real* left, Real* right);
bool is_neg(Real real);
int Twos(int op);
int main(int argc, char* argv[])
{
int left, right;
char op;
int value;
Real rLeft, rRight, result;
if (argc < 4) {
printf("Usage: %s <left> <op> <right>\n", argv[0]);
return -1;
}
sscanf(argv[1], "%f", (float*)&left);
sscanf(argv[2], "%c", &op);
sscanf(argv[3], "%f", (float*)&right);
rLeft = Decode(left);
rRight = Decode(right);
if (op == '+') {
result = Add(rLeft, rRight);
}
else {
printf("Unknown operator '%c'\n", op);
return -2;
}
value = Encode(result);
printf("%.3f %c %.3f = %.3f (0x%08x)\n",
*((float*)&left),
op,
*((float*)&right),
*((float*)&value),
value
);
return 0;
}
Real Decode(int float_value)
{ // Test sign bit of float_value - Test exponent bits of float_value & apply bias - Test mantissa bits of float_value
Real result{ float_value >> 31 & 1 ? 1 : 0, ((long)Add(float_value >> 23 & 0xFF, -BIAS32)), (unsigned long)float_value & 0x7FFFFF };
return result;
};
int Encode(Real real_value)
{
int x = 0;
x |= real_value.fraction; // Set the fraction bits of x
x |= real_value.sign << 31; // Set the sign bits of x
x |= Add(real_value.exponent, BIAS32) << 23; // Set the exponent bits of x
return x;
}
Real Normalize(Real value)
{
if (is_neg(value))
{
value.fraction = Twos(value.fraction);
}
unsigned int i = 0;
while (i < 9)
{
if ((value.fraction >> Add(23, i)) & 1) // If there are set bits past the mantissa section
{
value.fraction >>= 1; // shift mantissa right by 1
value.exponent = Add(value.exponent, 1); // increment exponent to accomodate for shift
}
i = Add(i, 1);
}
return value;
}
Real Add(Real left, Real right)
{
Real a = left, b = right;
alignExponents(&a, &b); // Aligns exponents of both operands
unsigned long sum = Add(a.fraction, b.fraction);
Real result = Normalize({ a.sign, a.exponent, sum }); // Normalize result if need be
return result;
}
unsigned long Add(unsigned long leftop, unsigned long rightop)
{
unsigned long sum = 0, test = 1; // sum initialized to 0, test created to compare bits
while (test) // while test is not 0
{
if (leftop & test) // if the digit being tested is 1
{
if (sum & test) sum ^= test << 1; // if the sum tests to 1, carry a bit over
sum ^= test;
}
if (rightop & test)
{
if (sum & test) sum ^= test << 1;
sum ^= test;
}
test <<= 1;
}
return sum;
}
void alignExponents(Real* a, Real* b)
{
if (a->exponent != b->exponent) // If the exponents are not equal
{
if (a->exponent > b->exponent)
{
int disp = a->exponent - b->exponent; // number of shifts needed based on difference between two exponents
b->fraction |= 1 << 23; // sets the implicit bit for shifting
b->exponent = a->exponent; // sets exponents equal to each other
b->fraction >>= disp; // mantissa is shifted over to accomodate for the increase in power
return;
}
int disp = b->exponent - a->exponent;
a->fraction |= 1 << 23;
a->exponent = b->exponent;
a->fraction >>= disp;
return;
}
return;
}
bool is_neg(Real real)
{
if (real.sign) return true;
return false;
}
int Twos(int op)
{
return Add(~op, -1); // NOT the operand and add 1 to it
}
On top of that, I just tested the values 10.5 + 5.5 and got a 24.0, so there appears to be even more wrong with this than I initially thought. I've been working on this for days and would love some help/advice.
Here is some help/advice. Now that you have worked on some of the code, I suggest going back and reworking your data structure. The declaration of such a crucial data structure would benefit from a lot more comments, making sure you know exactly what each field means.
For example, the implicit bit is not always 1. It is zero if the exponent is zero. That should be dealt with in your Encode and Decode functions. For the rest of your code, it is just a significand bit and should not have any special handling.
When you start thinking about rounding, you will find you often need more than 23 bits in an intermediate result.
Making the significand of negative numbers 2's complement will create a problem of having the same information stored two ways. You will have both a sign bit as though doing sign-and-magnitude and have the sign encoded in the signed integer signficand. Keeping them consistent will be a mess. Whatever you decide about how Real will store negative numbers, document it and keep it consistent throughout.
If I were implementing this I would start by defining Real very, very carefully. I would then decide what operations I wanted to be able to do on Real, and write functions to do them. If you get those right each function will be relatively simple.
I'm programming a class to handle integers with a bitset with more bits of a unsigned long long.
#include <bitset>
#include <string>
#define MAX_BITS 32000
class RossInt {
std::bitset<MAX_BITS> bits;
public:
RossInt(unsigned long long num); /* Copies the bits of num into the bitset */
RossInt operator+ (const RossInt& op) const;
std::string to_string() const; /* Returns the number in decimal in a string */
};
Since the number is bigger than a unsigned long long I'll put it in a string, but the issue is that I can't use the way I would normally use with the decimal_n += pow(2, bit_index) cause I can't store the result into a variable and I want to use as less external library as possible.
Is there a way for converting it using bitwise operators or any other way?
After loosing two nights on it, I've finally found a way of printing it. It's not the most efficient way and I'm gonna rewrite it until I'm satisfied with the speed.
std::ostream& operator<<(std::ostream& os, const RossInt& value) {
RossInt op = value;
RossInt i = 1;
while (op / i > 0) i = i * 10;
do {
i = i / 10;
os << (op / i).to_ullong();
op = op % i;
} while (i > 1);
return os;
}
The logic used in this code is pretty simple: if x = y and I divide both of them by 10^x, x is still equal to y. Using this property I wrote a code that prints the number digit by digit.
The first line copies the number to avoid changes in the value.
RossInt i = 1;
while (op / i > 0) i = i * 10;
The initial value of i is 1, the neutral number for multiplications. In the while it keeps increasing i until it gets bigger than the value of the number making possible to know the number of digit of the number without converting it in decimal.
do {
i = i / 10;
os << (op / i).to_ullong();
op = op % i;
} while (i > 1);
This cycle is used to do the real print of the number.
With every iteration the i loose a 0. Dividing op by i removes as many digits from the bottom of op as many 0s i has. Ex.: 12345 / 100 = 123.
The operation right after that does the exact opposite: it keeps as many digits as the 0s instead of removing them.
This cycle then print a digit and remove it right after, passing and printing them all.
What you are looking for is a Binary->BCD (Binary Coded Decimal) solution. Once you have a BCD, printing it is easy.
Given BCD math is heavily used I tried looking for a standard solution, but could not find one.
This example class should help. It's not traditional BCD representation, but it's easier to follow.
class BCD
{
public:
// function to make the "root" value.
static BCD One() { BCD retval; retval.digits = {1}; return retval; }
// we will generate all powers of two from doubling from one
void Double()
{
for(auto& d : digits) d *= 2;
Normalise();
}
// We can generate all other numbers by combining powers of two.
void operator+=(const BCD& other)
{
if (other.digits.size()>digits.size())
digits.resize(other.digits.size(), 0);
std::transform(digits.begin(), digits.end(), other.digits.begin(), digits.begin(), std::plus<char>{});
Normalise();
}
friend inline std::ostream& operator << (std::ostream&, const BCD&);
private:
// "normal form" is all digits 0-9, if required carry the one
void Normalise()
{
int carry = 0;
for(auto& d : digits) {
d+=carry;
carry = d/10;
d = d%10;
}
if (carry) digits.push_back(carry);
}
std::vector<char> digits;
};
inline std::ostream& operator<<(std::ostream& os, const BCD& bcd)
{
for_each(bcd.digits.rbegin(), bcd.digits.rend(), [&os](char d){ os << (char)(d+'0'); });
}
Using this class it should be simple to build a BCD from a bitset. For example
int main()
{
BCD power_of_two = BCD::One();
BCD total;
for (int i = 0; i < 10; ++i) {
total += power_of_two;
std::cout << power_of_two << " total = " << total << "\n";
power_of_two.Double();
}
return 0;
}
produces
1 total = 1
2 total = 3
4 total = 7
8 total = 15
16 total = 31
32 total = 63
64 total = 127
128 total = 255
256 total = 511
512 total = 1023
all you have to do is make the "+=" conditional on the bit in the set. I'm not claiming this is the best way to do it, but it should be good enough to test your results.
Hope it helps
-----------------Update after comments----------
It has been pointed out that my solution is not a complete solution.
In the spirit of being helpful I will provide how you can use the BCD class above to build a "to_string" function.
If you don't understand this you have to read up on number theory. You have to do it this way because 10=5*2. 5 is odd, and therefore there is no 10^x = 2^y, where x and y are integers. In practice, this means that the first digit in the decimal representation depends on every digit in the binary representation. You must determine it by adding all the powers of two. You may as well generate the whole number using the same logic.
Please be advised though that stack overflow exists to help programmers who have got stuck, not to do other peoples work for them. Please do not expect to have other people post working programs for you.
std::string to_string() const
{
BCD power_of_two = BCD::One();
BCD total;
for (int idx = 0; idx < bits.size(); ++idx) {
if (bits.test(idx))
total += power_of_two;
power_of_two.Double();
}
std::ostringstream os;
os << total;
return os.str();
}
I am currently using the code below that removes all digits equal to zero from an integer.
int removeZeros(int candid)
{
int output = 0;
string s(itoa(candid));
for (int i = s.size(); i != 0; --i)
{
if (s[i] != '0') output = output * 10 + atoi(s[i]);
}
return output;
}
The expected output for e.g. 102304 would be 1234.
Is there a more compact way of doing this by directly working on the integer, that is, not string representation? Is it actually going to be faster?
Here's a way to do it without strings and buffers.
I've only tested this with positive numbers. To make this work with negative numbers is an exercise left up to you.
int removeZeros(int x)
{
int result = 0;
int multiplier = 1;
while (x > 0)
{
int digit = x % 10;
if (digit != 0)
{
int val = digit * multiplier;
result += val;
multiplier *= 10;
}
x = x / 10;
}
return result;
}
For maintainability, I would suggest, don't work directly on the numeric value. You can express your requirements in a very straightforward way using string manipulations, and while it's true that it will likely perform slower than number manipulations, I expect either to be fast enough that you don't have to worry about the performance unless it's in an extremely tight loop.
int removeZeros(int n) {
auto s = std::to_string(n);
s.erase(std::remove(s.begin(), s.end(), '0'), s.end());
return std::stoi(s);
}
As a bonus, this simpler implementation handles negative numbers correctly. For zero, it throws std::invalid_argument, because removing all zeros from 0 doesn't produce a number.
You could try something like this:
template<typename T> T nozeros( T const & z )
{
return z==0 ? 0 : (z%10?10:1)*nozeros(z/10)+(z%10);
}
If you want to take your processing one step further you can do a nice tail recursion , no need for a helper function:
template<typename T> inline T pow10(T p, T res=1)
{
return p==0 ? res : pow10(--p,res*10);
}
template<typename T> T nozeros( T const & z , T const & r=0, T const & zp =0)
{
static int digit =-1;
return not ( z ^ r ) ? digit=-1, zp : nozeros(z/10,z%10, r ? r*pow10(++digit)+zp : zp);
}
Here is how this will work with input 32040
Ret, z, r, zp, digits
-,32040,0,0, -1
-,3204,0,0, -1
-,320,4,0,0, -1
-,32,0,4,4, 0
-,3,2,4, 0
-,0,3,24, 1
-,0,0,324, 2
324,-,-,-, -1
Integer calculations are always faster than actually transforming your integer to string, making comparisons on strings, and looking up strings to turn them back to integers.
The cool thing is that if you try to pass floats you get nice compile time errors.
I claim this to be slightly faster than other solutions as it makes less conditional evaluations which will make it behave better with CPU branch prediction.
int number = 9042100;
stringstream strm;
strm << number;
string str = strm.str();
str.erase(remove(str.begin(), str.end(), '0'), str.end());
number = atoi(str.c_str());
No string representation is used here. I can't say anything about the speed though.
int removezeroes(int candid)
{
int x, y = 0, n = 0;
// I did this to reverse the number as my next loop
// reverses the number while removing zeroes.
while (candid>0)
{
x = candid%10;
n = n *10 + x;
candid /=10;
}
candid = n;
while (candid>0)
{
x = candid%10;
if (x != 0)
y = y*10 + x;
candid /=10;
}
return y;
}
If C++11 is available, I do like this with lambda function:
int removeZeros(int candid){
std::string s=std::to_string(candid);
std::string output;
std::for_each(s.begin(), s.end(), [&](char& c){ if (c != '0') output += c;});
return std::stoi(output);
}
A fixed implementation of g24l recursive solution:
template<typename T> T nozeros(T const & z)
{
if (z == 0) return 0;
if (z % 10 == 0) return nozeros(z / 10);
else return (z % 10) + ( nozeros(z / 10) * 10);
}
I have implemented class NaturalNum for representing a natural number of "infinite" size (up to 4GB).
I have also implemented class RationalNum for representing a rational number with infinite accuracy. It stores the numerator and the denominator of the rational number, both of which are NaturalNum instances, and relies on them when performing any arithmetic operation issued by the user.
The only place where precision is "dropped by a certain degree", is upon printing, since there's a limit (provided by the user) to the number of digits that appear after the decimal (or non-decimal) point.
My question concerns one of the constructors of class RationalNum. Namely, the constructor that takes a double value, and computes the corresponding numerator and denominator.
My code is given below, and I would like to know if anyone sees a more accurate way for computing them:
RationalNum::RationalNum(double value)
{
if (value == value+1)
throw "Infinite Value";
if (value != value)
throw "Undefined Value";
m_sign = false;
m_numerator = 0;
m_denominator = 1;
if (value < 0)
{
m_sign = true;
value = -value;
}
// Here is the actual computation
while (value > 0)
{
unsigned int floor = (unsigned int)value;
value -= floor;
m_numerator += floor;
value *= 2;
m_numerator *= 2;
m_denominator *= 2;
}
NaturalNum gcd = GCD(m_numerator,m_denominator);
m_numerator /= gcd;
m_denominator /= gcd;
}
Note: variables starting with 'm_' are member variables.
Thanks
The standard library contains a function for obtaining the significand and exponent, frexp.
Just multiply the significand to get all bits before decimal point and set appropriate denominator. Just don't forget the significand is normalized to be between 0.5 and 1 (I would consider between 1 and 2 more natural but whatever) and that it has 53 significant bits for IEEE double (there are no practically used platforms that would use different floating point format).
I'm not 100% confident in the math that you have for the actual computation only because I haven't really examined it, but I think the below method removes the need to use the GCD function which could bring in some unnecessary running time.
Here is the class I came up with. I haven't fully tested it, but I produced a couple billion random doubles and the asserts never fired, so I'm reasonably confident in its usability, but I would still test the edge cases around INT64_MAX a little more.
If I'm not mistaken, the running time complexity of this algorithm is linear with respect to the size in bits of the input.
#include <iostream>
#include <cmath>
#include <cassert>
#include <limits>
class Real;
namespace std {
inline bool isnan(const Real& r);
inline bool isinf(const Real& r);
}
class Real {
public:
Real(double val)
: _val(val)
{
if (std::isnan(val)) { return; }
if (std::isinf(val)) { return; }
double d;
if (modf(val, &d) == 0) {
// already a whole number
_num = val;
_den = 1.0;
return;
}
int exponent;
double significand = frexp(val, &exponent); // val = significand * 2^exponent
double numerator = val;
double denominator = 1;
// 0.5 <= significand < 1.0
// significand is a fraction, multiply it by two until it's a whole number
// subtract exponent appropriately to maintain val = significand * 2^exponent
do {
significand *= 2;
--exponent;
assert(std::ldexp(significand, exponent) == val);
} while (modf(significand, &d) != 0);
assert(exponent <= 0);
// significand is now a whole number
_num = significand;
_den = 1.0 / std::ldexp(1.0, exponent);
assert(_val == _num / _den);
}
friend std::ostream& operator<<(std::ostream &os, const Real& rhs);
friend bool std::isnan(const Real& r);
friend bool std::isinf(const Real& r);
private:
double _val = 0;
double _num = 0;
double _den = 0;
};
std::ostream& operator<<(std::ostream &os, const Real& rhs) {
if (std::isnan(rhs) || std::isinf(rhs)) {
return os << rhs._val;
}
if (rhs._den == 1.0) {
return os << rhs._num;
}
return os << rhs._num << " / " << rhs._den;
}
namespace std {
inline bool isnan(const Real& r) { return std::isnan(r._val); }
inline bool isinf(const Real& r) { return std::isinf(r._val); }
}
#include <iomanip>
int main () {
#define PRINT_REAL(num) \
std::cout << std::setprecision(100) << #num << " = " << num << " = " << Real(num) << std::endl
PRINT_REAL(1.5);
PRINT_REAL(123.875);
PRINT_REAL(0.125);
// double precision issues
PRINT_REAL(-10000000000000023.219238745);
PRINT_REAL(-100000000000000000000000000000000000000000.5);
return 0;
}
Upon looking at your code a little bit more, there's at least a problem with your testing for infinite values. Note the following program:
#include <numeric>
#include <cassert>
#include <cmath>
int main() {
{
double d = std::numeric_limits<double>::max(); // about 1.7976931348623e+308
assert(!std::isnan(d));
assert(!std::isinf(d));
// assert(d != d + 1); // fires
}
{
double d = std::ldexp(1.0, 500); // 2 ^ 700
assert(!std::isnan(d));
assert(!std::isinf(d));
// assert(d != d + 1); // fires
}
}
In addition to that, if your GCD function doesn't support doubles, then you'll be limiting yourself in terms of values you can import as doubles. Try any number > INT64_MAX and the GCD function may not work.
Can anybody give an example of c++ code that can easily convert a decimal value to binary and a binary value to decimal please?
Well, your question is really vague, so this answer is the same.
string DecToBin(int number)
{
if ( number == 0 ) return "0";
if ( number == 1 ) return "1";
if ( number % 2 == 0 )
return DecToBin(number / 2) + "0";
else
return DecToBin(number / 2) + "1";
}
int BinToDec(string number)
{
int result = 0, pow = 1;
for ( int i = number.length() - 1; i >= 0; --i, pow <<= 1 )
result += (number[i] - '0') * pow;
return result;
}
You should check for overflow and do input validation of course.
x << 1 == x * 2
Here's a way to convert to binary that uses a more "programming-like" approach rather than a "math-like" approach, for lack of a better description (the two are actually identical though, since this one just replaces divisions by right shifts, modulo by a bitwise and, recursion with a loop. It's kind of another way of thinking about it though, since this makes it obvious you are extracting the individual bits).
string DecToBin2(int number)
{
string result = "";
do
{
if ( (number & 1) == 0 )
result += "0";
else
result += "1";
number >>= 1;
} while ( number );
reverse(result.begin(), result.end());
return result;
}
And here is how to do the conversion on paper:
Decimal to binary
Binary to decimal
strtol will convert a binary string like "011101" to an internal value (which will normally be stored in binary as well, but you don't need to worry much about that). A normal conversion (e.g. operator<< with std:cout) will give the same value in decimal.
//The shortest solution to convert dec to bin in c++
void dec2bin(int a) {
if(a!=0) dec2bin(a/2);
if(a!=0) cout<<a%2;
}
int main() {
int a;
cout<<"Enter the number: "<<endl;
cin>>a;
dec2bin(a);
return 0;
}
I assume you want a string to binary conversion?
template<typename T> T stringTo( const std::string& s )
{
std::istringstream iss(s);
T x;
iss >> x;
return x;
};
template<typename T> inline std::string toString( const T& x )
{
std::ostringstream o;
o << x;
return o.str();
}
use these like this:
int x = 32;
std:string decimal = toString<int>(x);
int y = stringTo<int>(decimal);