Self Made Pow() C++

How can I write a power function myself?

Negative powers are not a problem, they're just the inverse (1/x) of the positive power.

Floating point powers are just a little bit more complicated; as you know a fractional power is equivalent to a root (e.g. x^(1/2) == sqrt(x)) and you also know that multiplying powers with the same base is equivalent to add their exponents.

With all the above, you can:

Decompose the exponent in a integer part and a rational part.
Calculate the integer power with a loop (you can optimise it decomposing in factors and reusing partial calculations).
Calculate the root with any algorithm you like (any iterative approximation like bisection or Newton method could work).
Multiply the result.
If the exponent was negative, apply the inverse.

Example:

2^(-3.5) = (2^3 * 2^(1/2)))^-1 = 1 / (2*2*2 * sqrt(2))

The most efficient way to implement an integer based power function pow(int, int)

Exponentiation by squaring.

int ipow(int base, int exp)
{
    int result = 1;
    for (;;)
    {
        if (exp & 1)
            result *= base;
        exp >>= 1;
        if (!exp)
            break;
        base *= base;
    }

    return result;
}

This is the standard method for doing modular exponentiation for huge numbers in asymmetric cryptography.

How is pow() calculated in C?

If you're curious how the pow function might be implemented in practice, you can look at the source code. There is a kind of "knack" to searching through unfamiliar (and large) codebases to find the section you are looking for, and it's good to get some practice.

One implementation of the C library is glibc, which has mirrors on GitHub. I didn't find an official mirror, but an unofficial mirror is at https://github.com/lattera/glibc

We first look at the math/w_pow.c file which has a promising name. It contains a function __pow which calls __ieee754_pow, which we can find in sysdeps/ieee754/dbl-64/e_pow.c (remember that not all systems are IEEE-754, so it makes sense that the IEEE-754 math code is in its own directory).

It starts with a few special cases:

if (y == 1.0) return x;
if (y == 2.0) return x*x;
if (y == -1.0) return 1.0/x;
if (y == 0) return 1.0;

A little farther down you find a branch with a comment

/* if x<0 */

Which leads us to

return (k==1)?__ieee754_pow(-x,y):-__ieee754_pow(-x,y); /* if y even or odd */

So you can see, for negative x and integer y, the glibc version of pow will compute pow(-x,y) and then make the result negative if y is odd.

This is not the only way to do things, but my guess is that this is common to many implementations. You can see that pow is full of special cases. This is common in library math functions, which are supposed to work correctly with unfriendly inputs like denormals and infinity.

The pow function is especially hard to read because it is heavily-optimized code which does bit-twiddling on floating-point numbers.

The C Standard

The C standard (n1548 §7.12.7.4) has this to say about pow:

A domain error occurs if x is finite and negative and y is finite and not an integer value.

So, according to the C standard, negative x should work.

There is also the matter of appendix F, which gives much tighter constraints on how pow works on IEEE-754 / IEC-60559 systems.

Write Pow Function Without math.h in C

Everything seems perfect except for one condition :- when b<0.

For b<0,simply return

return (1.0/a)*MyPow(a,abs(b)-1);   //where  abs(b) is  absolute value of b.

return (1.0/a)*(MyPow(a,b+1));

Also,your definition of function is not valid for performing negative exponentiation,you should change it to

float MyPow(int a,int b)

how could I use the power function in c/c++ without pow(), functions, or recursion

It is a series. Replace pow() based on the previous iteration. @Bathsheba

Code does not need to call pow(). It can form pow(x, 5 * i - 1) and pow(-1, i - 1), since both have an int exponent based on the iterator i, from the prior loop iteration.

Example:

Let f(x, i) = pow(x, 5 * i - 1)

Then f(x, 1) = x*x*x*x

and f(x, i > 1) = f(x, i-1) * x*x*x*x*x

double power_n1 = 1.0; 
double power_x5 = x*x*x*x; 
for (int i = 1; i < j + 1; i++) 
  // sum += (pow(-1, i - 1)) / (5 * i - 1) * (pow(x, 5 * i - 1));
  sum += power_n1 / (5 * i - 1) * power_x5;
  power_n1 = -power_n1;
  power_x5 *= x*x*x*x*x; 
}

Does pow() work for int data type in C?

Floating point precision is doing its job here. The actual working of pow is using log

pow(a, 2) ==> exp(log(a) * 2)

Look at math.h library which says:
###<math.h>

/* Excess precision when using a 64-bit mantissa for FPU math ops can
cause unexpected results with some of the MSVCRT math functions. For
example, unless the function return value is stored (truncating to
53-bit mantissa), calls to pow with both x and y as integral values
sometimes produce a non-integral result. ... */

Just add 0.5 to the return value of pow and then convert it to int.

b = (int)(pow(a,2) + 0.5);

So, the answer to your question

Does pow() work for int data type in C?

Not always. For integer exponentiation you could implement your own function (this will work for 0 and +ve exp only):

unsigned uint_pow(unsigned base, unsigned exp)
{
    unsigned result = 1;
    while (exp)
    {
        if (exp % 2)
           result *= base;
        exp /= 2;
        base *= base;
    }
    return result;
}

C - Writing a function for an exponent without using Pow

Try this:

long int x_to_the_n (int x,int n)
{
    int i; /* Variable used in loop counter */
    int number = 1;

    for (i = 0; i < n; ++i)
        number *= x;

    return(number);
}

C++ program for power but without using pow function

The program must not use any pre-defined C++ functions (like pow function) for this task

You can use some piece of c++ code like follows, to compute x^y, without using any predefined function:

int x = 5;
int y = 3;
int result = 1;
for(int i = 0; i < y; ++i)
{
    result *= x;
}

cout << result << endl;

Output:

See a working sample here.

Pow function in C is outputting zero

More than likely, your problem is a common error when dealing with numbers in C. Consider this:

double recN = 1/N;

Here, you define the variable as double, but you then perform integer division. If N>1 then this will result in zero.

Instead, look for places you've done this type of calculation and convert it to:

double recN = 1.0/N;