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Given two integers X and Y, whats the most efficient way of converting them into X.Y float value in C++?
E.g.
X = 3, Y = 1415 -> 3.1415
X = 2, Y = 12 -> 2.12
Here are some cocktail-napkin benchmark results, on my machine, for all solutions converting two ints to a float, as of the time of writing.
Caveat: I've now added a solution of my own, which seems to do well, and am therefore biased! Please double-check my results.
Test
Iterations
ns / iteration
#aliberro's conversion v2
79,113,375
13
#3Dave's conversion
84,091,005
12
#einpoklum's conversion
1,966,008,981
0
#Ripi2's conversion
47,374,058
21
#TarekDakhran's conversion
1,960,763,847
0
CPU: Quad Core Intel Core i5-7600K speed/min/max: 4000/800/4200 MHz
Devuan GNU/Linux 3
Kernel: 5.2.0-3-amd64 x86_64
GCC 9.2.1, with flags: -O3 -march=native -mtune=native
Benchmark code (Github Gist).
float sum = x + y / pow(10,floor(log10(y)+1));
log10 returns log (base 10) of its argument. For 1234, that'll be 3 point something.
Breaking this down:
log10(1234) = 3.091315159697223
floor(log10(1234)+1) = 4
pow(10,4) = 10000.0
3 + 1234 / 10000.0 = 3.1234.
But, as #einpoklum pointed out, log(0) is NaN, so you have to check for that.
#include <iostream>
#include <cmath>
#include <vector>
using namespace std;
float foo(int x, unsigned int y)
{
if (0==y)
return x;
float den = pow(10,-1 * floor(log10(y)+1));
return x + y * den;
}
int main()
{
vector<vector<int>> tests
{
{3,1234},
{1,1000},
{2,12},
{0,0},
{9,1}
};
for(auto& test: tests)
{
cout << "Test: " << test[0] << "," << test[1] << ": " << foo(test[0],test[1]) << endl;
}
return 0;
}
See runnable version at:
https://onlinegdb.com/rkaYiDcPI
With test output:
Test: 3,1234: 3.1234
Test: 1,1000: 1.1
Test: 2,12: 2.12
Test: 0,0: 0
Test: 9,1: 9.1
Edit
Small modification to remove division operation.
(reworked solution)
Initially, my thoughts were improving on the performance of power-of-10 and division-by-power-of-10 by writing specialized versions of these functions, for integers. Then there was #TarekDakhran's comment about doing the same for counting the number of digits. And then I realized: That's essentially doing the same thing twice... so let's just integrate everything. This will, specifically, allow us to completely avoid any divisions or inversions at runtime:
inline float convert(int x, int y) {
float fy (y);
if (y == 0) { return float(x); }
if (y >= 1e9) { return float(x + fy * 1e-10f); }
if (y >= 1e8) { return float(x + fy * 1e-9f); }
if (y >= 1e7) { return float(x + fy * 1e-8f); }
if (y >= 1e6) { return float(x + fy * 1e-7f); }
if (y >= 1e5) { return float(x + fy * 1e-6f); }
if (y >= 1e4) { return float(x + fy * 1e-5f); }
if (y >= 1e3) { return float(x + fy * 1e-4f); }
if (y >= 1e2) { return float(x + fy * 1e-3f); }
if (y >= 1e1) { return float(x + fy * 1e-2f); }
return float(x + fy * 1e-1f);
}
Additional notes:
This will work for y == 0; but - not for negative x or y values. Adapting it for negative value is pretty easy and not very expensive though.
Not sure if this is absolutely optimal. Perhaps a binary-search for the number of digits of y would work better?
A loop would make the code look nicer; but the compiler would need to unroll it. Would it unroll the loop and compute all those floats beforehand? I'm not sure.
I put some effort into optimizing my previous answer and ended up with this.
inline uint32_t digits_10(uint32_t x) {
return 1u
+ (x >= 10u)
+ (x >= 100u)
+ (x >= 1000u)
+ (x >= 10000u)
+ (x >= 100000u)
+ (x >= 1000000u)
+ (x >= 10000000u)
+ (x >= 100000000u)
+ (x >= 1000000000u)
;
}
inline uint64_t pow_10(uint32_t exp) {
uint64_t res = 1;
while(exp--) {
res *= 10u;
}
return res;
}
inline double fast_zip(uint32_t x, uint32_t y) {
return x + static_cast<double>(y) / pow_10(digits_10(y));
}
double IntsToDbl(int ipart, int decpart)
{
//The decimal part:
double dp = (double) decpart;
while (dp > 1)
{
dp /= 10;
}
//Joint boths parts
return ipart + dp;
}
Simple and very fast solution is converting both values x and y to string, then concatenate them, then casting the result into a floating number as following:
#include <string>
#include <iostream>
std::string x_string = std::to_string(x);
std::string y_string = std::to_string(y);
std::cout << x_string +"."+ y_string ; // the result, cast it to float if needed
(Answer based on the fact that OP has not indicated what they want to use the float for.)
The fastest (most efficient) way is to do it implicitly, but not actually do anything (after compiler optimizations).
That is, write a "pseudo-float" class, whose members are integers of x and y's types before and after the decimal point; and have operators for doing whatever it is you were going to do with the float: operator+, operator*, operator/, operator- and maybe even implementations of pow(), log2(), log10() and so on.
Unless what you were planning to do is literally save a 4-byte float somewhere for later use, it would almost certainly be faster if you had the next operand you need to work with then to really create a float from just x and y, already losing precision and wasting time.
Try this
#include <iostream>
#include <math.h>
using namespace std;
float int2Float(int integer,int decimal)
{
float sign = integer/abs(integer);
float tm = abs(integer), tm2 = abs(decimal);
int base = decimal == 0 ? -1 : log10(decimal);
tm2/=pow(10,base+1);
return (tm+tm2)*sign;
}
int main()
{
int x,y;
cin >>x >>y;
cout << int2Float(x,y);
return 0;
}
version 2, try this out
#include <iostream>
#include <cmath>
using namespace std;
float getPlaces(int x)
{
unsigned char p=0;
while(x!=0)
{
x/=10;
p++;
}
float pow10[] = {1.0f,10.0f,100.0f,1000.0f,10000.0f,100000.0f};//don't need more
return pow10[p];
}
float int2Float(int x,int y)
{
if(y == 0) return x;
float sign = x != 0 ? x/abs(x) : 1;
float tm = abs(x), tm2 = abs(y);
tm2/=getPlaces(y);
return (tm+tm2)*sign;
}
int main()
{
int x,y;
cin >>x >>y;
cout << int2Float(x,y);
return 0;
}
If you want something that is simple to read and follow, you could try something like this:
float convertToDecimal(int x)
{
float y = (float) x;
while( y > 1 ){
y = y / 10;
}
return y;
}
float convertToDecimal(int x, int y)
{
return (float) x + convertToDecimal(y);
}
This simply reduces one integer to the first floating point less than 1 and adds it to the other one.
This does become a problem if you ever want to use a number like 1.0012 to be represented as 2 integers. But that isn't part of the question. To solve it, I would use a third integer representation to be the negative power of 10 for multiplying the second number. IE 1.0012 would be 1, 12, 4. This would then be coded as follows:
float convertToDecimal(int num, int e)
{
return ((float) num) / pow(10, e);
}
float convertToDecimal(int x, int y, int e)
{
return = (float) x + convertToDecimal(y, e);
}
It a little more concise with this answer, but it doesn't help to answer your question. It might help show a problem with using only 2 integers if you stick with that data model.
Say I choose a value for x that can be between 0 and 2147483647. (Int32.MaxValue)
I am trying to figure out how I can find values for a,b,n so that (a^b)%n=x
I already know that I can use ModPow to verify the values, but I don't know how I can find a fitting a,b and n.
#include <iostream>
/// Calculate (a^b)%n
/// \param a The base
/// \param b The exponent
/// \param n The modulo
/// \return (a^b)%n
int ModPow(int a, int b, int n) {
long long x = 1, y = a;
while (b > 0) {
if (b % 2 == 1) {
x = (x * y) % n; // multiplying with base
}
y = (y * y) % n; // squaring the base
b /= 2;
}
return x % n;
}
int main() {
int x = 1337;
// How to find a,b,n so that (a^b)%n=x
int a = ?;
int b = ?;
int n = ?;
if(x == ModPow(a,b,n))
printf("ok");
return 0;
}
int n = 2147483647
int a = ModPow(x, 9241, n);
int b = 464773;
n = 231 − 1 is a prime number. So due to Fermat's little theorem, xn mod n = x and xn − 1 mod n = 1 (unless x = 0) so x2 n − 1 mod n = x, too. 2 n − 1 = 9241 × 464773. So (x9241 mod n)464773 mod n = x. Note that you need x < n for this to work; x = 2147483647 cannot work if n is a 31 bit (i.e. signed) integer, too.
It took me a while to get here; for a long time I've had this answer messing about with Carmichael numbers and the Carmichael function before I reached this easy solution. See edit history for details.
The modulus operator:
Yields the remainder given by the following expression, where e1 is the first operand and e2 is the second: e1 – (e1 / e2) * e2
Therefor whatever the max value of x is, n must be larger. Since you're validating with n as an int and you're specifying the range: 0 and numeric_limits<int>::max() that must be an exclusive range, and for n to be an int the only possible value it can take is: numeric_limits<int>::max().
With n forced our equation effectively becomes: ab = x.
We'll need to do a check here that x is not 1, if it is b = 0 and a can be anything in our legal range so we can arbitrarily pick a = 2. But baring this:
Our requirements are:
1 < a < x and a is an int
1 < b < x and b is an int
Given x, we can search for a combination of a and b that will fit as follows:
auto a = 0.0;
auto b = 1;
if(x == 1) {
a = 2.0;
b = 0;
} else {
while((a = pow(x, 1.0 / ++b)) > 2.0) {
double dummy;
if(modf(a, &dummy) == 0.0) {
break;
}
}
}
At this point, if a >= 2.0 then there is a valid solution to the problem. Now as you as you are probably well aware, pow is a very expensive function so this will likely take a very long time to execute for larger values of x, I'd personally suggest finding an a and b for every number for which such a pair exists and storing them in a map and doing a lookup on that.
Anyway this is a demonstration of working code:
Live Example
I want to compute the distance between numbers with help of the system described in the attached image.
For example: distance between 7 and 5 is -2, distance between 7 and 1 is 2 etc...
Any ideas how to do this in c++? The prefered direction is counter clockwise...
I am using a (int) vector.
If you do it in straightforward way (by considering all possibilities) it might look as follows
int distance(int a, int b)
{ // Distance from `a` to `b`
int d = b - a;
return
a <= b ?
(d <= +4 ? d : d - 8) :
(d <= -4 ? d + 8 : d);
}
which, if you prefer, can be rewritten as
int distance(int a, int b)
{ // Distance from `a` to `b`
int d = b - a;
return -4 < d && d <= 4 ? d : (d > 0 ? d - 8 : d + 8);
}
An alternative, more elegant approach would be to always calculate the positive CCW distance and flip it to negative CW distance if it is greater than 4
int distance(int a, int b)
{ // Distance from `a` to `b`
int d = (b + 8 - a) % 8;
// `d` is CCW distance from `a` to `b`
return d <= 4 ? d : d - 8;
}
But if you want the compiler to generate the most efficient code for this, follow the golden rule "use unsigned types everywhere you can, use signed types only if you have to":
int distance(unsigned a, unsigned b)
{ // Distance from `a` to `b`
unsigned d = (b + 8 - a) % 8;
// `d` is CCW distance from `a` to `b`
return d <= 4 ? d : (int) d - 8;
}
These are really complicated answers. Here is a simpler one:
int distance(int x, int y) {
int d = (y - x) & 7;
return d > 4 ? d - 8 : d;
}
This always returns a result in the range -3..+4. Modular arithmetic is a little simpler to write when the ring size is a power of two, as is the case here.
distance(7, 5) = -2
distance(5, 7) = +2
distance(6, 2) = +4
distance(2, 6) = +4
We use & 7 because it is the simplest way to get the modulo. Alternatively, you can use % 8, but you must also add 8 in order to make sure that the input is not negative:
int d = (y - x + 8) % 8; // same result
Alternatively, you can handle negative numbers explicitly:
int d = (y - x) % 8;
if (d < 0) {
d += 8;
}
// same result
This is just a matter of style.
For simplicity you can find the element from std::find and get the distance from start from std::distance
for example
as you mentioned the data saved in int vector
std::vector<int>::iterator it1 = std::find(myvec.begin(), myvec.end(), val_1);
std::vector<int>::iterator it2 = std::find(myvec.begin(), myvec.end(), val_2);
int dist = std::distance(myvec.begin(),it1) - std::distance(myvec.begin.it2);
if(dist < 0) return dist
else(dist > 0) return myvector.size() - dist()
So hope this will give the distance as the image ...
I am pretty sure this works:
list = [0,1,2,3,4,5,6,7]
distance(x,y) {
a = y-x
b = length(list)-abs(y-x)
z = min(abs(a), abs(b))
if(z=abs(a)) { return a }
if(z=abs(b)) { return b }
}
where abs() is the mathematical absolute value function.
I make a few assumptions here.
As #Hédi Ghédiri pointed out, you are not counting counter-clockwise both times. I am assuming you count the shortest path to the number. (I used the mathematical min() function)
You prefer the positive value over the negative value (#Harper's comment). If you prefer the negative value, switch the last two if statements.
There may be a more concise method, but this (hopefully) works. Please comment if it is wrong. Hope this is helpful!
Edit: this is psuedocode. It should be easy to write in c++. Use the abs() function in <stdlib.h> Forget about list and length(list). Use int types for the variables, and everything else should work.
The following code is prepared to meet all of your needs, for example I assume, that if direction is clockwise the distance is to be negative.
#include <iostream>
#define RING_SIZE 8
enum direction
{
clockwise,
counterClockwise
};
int distance(int a, int b, direction dir)
{
int dist;
if(dir == clockwise)
{
if(a>b)
{
dist = -(a-b);
}
else
{
dist =-(RING_SIZE-b+a);
}
}
else
{
if(a<b)
{
dist = b-a;
}
else
{
dist = RING_SIZE-a+b;
}
}
if(a==b) dist = 0;//Add this if distance between same point must to be 0
return dist;
}
int main()
{
std::cout << distance(7, 2, clockwise) << std::endl;
}
I think this should work
int func(a,b)
{
dist=(b-a);
if(dist<0)
dist +=8;
return dist;
}
in case you're really stuck
Write the definition of a function minMax that has five parameters . The first three parameters are integers . The last two are set by the function to the largest and smallest of the values of the first three parameters . The function does not return a value .
The function can be used as follows:
int a=31, b=5, c=19 big, small; minMax(a,b,c,&big,&small); /* big is now 31 / / small is now 5 */
void minMax (int a, int b, int c, int *big, int *small) {
if (a > b && a > c)
*big = a;
else if (b>a && b>c)
*big = b;
else if (c>a && c>b)
*big = c;
if (a < b && a < c)
*small = a;
else if (b<a && b<c)
*small = b;
else if (c<a && c<b)
*small = c;
return;
}
Remarks:
⇒ Your function did not change the value of small. Make sure you are dereferencing it in your function.
Common Errors:
Make sure your if/else statements are correct.
Make sure you have semicolons at the end of each statement.
Make sure you are assigning the correct values to big and small.
If the 3 numbers a, b and c are all different, your code should work fine. But if 2 (or even all 3) values are the same, you will get an error, because you are always checking for "strictly greater than" and "strictly less than", that is, you are using > and < instead of >= and <=. Try it with a=31, B=31, C=31 and you will see that you will neither assign a value to *big nor to *small.
How about this:
void minMax(int x, int y, int z, int * big, int * small)
{
*big = x;
if (y > *big)
{
*big = y;
}
if (z > *big)
{
*big = z;
}
*small = x;
if (y < *small)
{
*small = y;
}
if (z < *small)
{
*small = z;
}
}
I managed to get my sqrt function to run perfectly, but I'm second guessing if I wrote this code correctly based on the pseudo code I was given.
Here is the pseudo code:
x = 1
repeat 10 times: x = (x + n / x) / 2
return x.
The code I wrote,
#include <iostream>
#include <math.h>
using namespace std;
double my_sqrt_1(double n)
{
double x= 1; x<10; ++x;
return (x+n/x)/2;
}
No, your code is not following your pseudo-code. For example, you're not repeating anything in your code. You need to add a loop to do that:
#include <iostream>
#include <math.h>
using namespace std;
double my_sqrt_1(double n)
{
double x = 1;
for(int i = 0; i < 10; ++i) // repeat 10 times
x = (x+n/x)/2;
return x;
}
Let's analyze your code:
double x = 1;
// Ok, x set to 1
x < 10;
// This is true, as 1 is less than 10, but it is not used anywhere
++x;
// Increment x - now x == 2
return (x + n / x) / 2
// return value is always (2 + n / 2) / 2
As you don't have any loop, function will always exit in the first "iteration" with the return value (2 + n / 2) / 2.
Just as another approach that you can use binary search or the another pretty elegant solution is to use the Newton's method.
Newton's method is a method for finding roots of a function, making use of a function's derivative. At each step, a value is calculated as: x(step) = x(step-1) - f(x(step-1))/f'(x(step-1)) Newton's_method
This might be faster than binary search.My implementation in C++:
double NewtonMethod(double x) {
double eps = 0.0001; //the precision
double x0 = 10;
while( fabs(x-x0) > eps) {
double a = x0*x0-n;
double r = a/(2*x0);
x = x0 - r;
x0 = x;
}
return x;
}
Since people are showing different approaches to calculating the square root, I couldn't resist ;)...
Below is the exact copy (with the original comments, but without preprocessor directives) of the inverse square root implementation from Quake III Arena:
float Q_rsqrt( float number )
{
long i;
float x2, y;
const float threehalfs = 1.5F;
x2 = number * 0.5F;
y = number;
i = * ( long * ) &y; // evil floating point bit level hacking
i = 0x5f3759df - ( i >> 1 ); // what the...?
y = * ( float * ) &i;
y = y * ( threehalfs - ( x2 * y * y ) ); // 1st iteration
// y = y * ( threehalfs - ( x2 * y * y ) ); // 2nd iteration, this can be removed
return y;
}