这是一个非常简洁的挑战。

编写一个带有两个参数的函数或过程，x并使用循环或内置的幂函数y返回WITHOUT 的结果。xy

获胜者是最具创造力的解决方案，将在3天后根据最高投票数选出。

杀伤人员地雷（7）

{×/⍵/⍺}

左参数为基，右参数为指数，例如：

     5 {×/⍵/⍺} 6
15625

说明：

• ⍵/⍺复制⍺ ⍵时间，例如5 {⍵/⍺} 6->5 5 5 5 5 5
• ×/拿产品，例如×/5 5 5 5 5 5-> 5×5×5×5×5×5->15625

C＃：浮点指数

好的，此解决方案非常脆弱。您可以通过向其抛出可笑的巨大数字（例如6）来轻松打破它。但是它对于诸如DoublePower(1.5, 3.4)，并且不使用递归！

    static double IntPower(double x, int y)
    {
        return Enumerable.Repeat(x, y).Aggregate((product, next) => product * next);
    }

    static double Factorial(int x)
    {
        return Enumerable.Range(1, x).Aggregate<int, double>(1.0, (factorial, next) => factorial * next);
    }

    static double Exp(double x)
    {
        return Enumerable.Range(1, 100).
            Aggregate<int, double>(1.0, (sum, next) => sum + IntPower(x, next) / Factorial(next));
    }

    static double Log(double x)
    {
        if (x > -1.0 && x < 1.0)
        {
            return Enumerable.Range(1, 100).
                Aggregate<int, double>(0.0, (sum, next) =>
                    sum + ((next % 2 == 0 ? -1.0 : 1.0) / next * IntPower(x - 1.0, next)));
        }
        else
        {
            return Enumerable.Range(1, 100).
                Aggregate<int, double>(0.0, (sum, next) =>
                    sum + 1.0 / next * IntPower((x - 1) / x, next));
        }
    } 

    static double DoublePower(double x, double y)
    {
        return Exp(y * Log(x));
    } 

C ++

模板元编程怎么样？它弯弯曲曲了一些小规则，但是值得一试：

#include <iostream>


template <int pow>
class tmp_pow {
public:
    constexpr tmp_pow(float base) :
        value(base * tmp_pow<pow-1>(base).value)
    {
    }
    const float value;
};

template <>
class tmp_pow<0> {
public:
    constexpr tmp_pow(float base) :
        value(1)
    {
    }
    const float value;
};

int main(void)
{
    tmp_pow<5> power_thirst(2.0f);
    std::cout << power_thirst.value << std::endl;
    return 0;
}

蟒蛇

def power(x,y):
    return eval(((str(x)+"*")*y)[:-1])

对于非整数幂不起作用。

Haskell-25个字符

f _ 0=1
f x y=x*f x (y-1)

遵循Marinus的APL版本：

f x y = product $ take y $ repeat x

移除mniip的注释和空格后，共有27个字符：

f x y=product$replicate y x

蟒蛇

如果y为正整数

def P(x,y):
    return reduce(lambda a,b:a*b,[x]*y)

JavaScript (ES6), 31

// Testable in Firefox 28
f=(x,y)=>eval('x*'.repeat(y)+1)

Usage:

> f(2, 0)
1
> f(2, 16)
65536

Explanation:

The above function builds an expression which multiply x y times then evaluates it.

I'm surprised to see that nobody wrote a solution with the Y Combinator, yet... thus:

Python2

Y = lambda f: (lambda x: x(x))(lambda y: f(lambda v: y(y)(v)))
pow = Y(lambda r: lambda (n,c): 1 if not c else n*r((n, c-1)))

No loops, No vector/list operations and No (explicit) recursion!

>>> pow((2,0))
1
>>> pow((2,3))
8
>>> pow((3,3))
27

C# : 45

Works for integers only:

int P(int x,int y){return y==1?x:x*P(x,y-1);}

bash & sed

No numbers, no loops, just an embarrasingly dangerous glob abuse. Preferably run in an empty directory to be safe. Shell script:

#!/bin/bash
rm -f xxxxx*
eval touch $(printf xxxxx%$2s | sed "s/ /{1..$1}/g")
ls xxxxx* | wc -l
rm -f xxxxx*

Javascript

function f(x,y){return ("1"+Array(y+1)).match(/[\,1]/g).reduce(function(l,c){return l*x;});}

Uses regular expressions to create an array of size y+1 whose first element is 1. Then, reduce the array with multiplication to compute power. When y=0, the result is the first element of the array, which is 1.

Admittedly, my goal was i) not use recursion, ii) make it obscure.

Mathematica

f[x_, y_] := Root[x, 1/y]

Probably cheating to use the fact that x^(1/y) = y√x

JavaScript

function f(x,y){return y--?x*f(x,y):1;}

Golfscript, 8 characters (including I/O)

~])*{*}*

Explanation:

TLDR: another "product of repeated array" solution.

The expected input is two numbers, e.g. 2 5. The stack starts with one item, the string "2 5".

Code     - Explanation                                             - stack
                                                                   - "2 5"
~        - pop "2 5" and eval into the integers 2 5                - 2 5        
]        - put all elements on stack into an array                 - [2 5]
)        - uncons from the right                                   - [2] 5
*        - repeat array                                            - [2 2 2 2 2]
{*}      - create a block that multiplies two elements             - [2 2 2 2 2] {*}
*        - fold the array using the block                          - 32

Ruby

class Symbol
  define_method(:**) {|x| eval x }
end

p(:****[$*[0]].*(:****$*[1]).*('*'))

Sample use:

$ ruby exp.rb 5 3
125
$ ruby exp.rb 0.5 3
0.125

This ultimately is the same as several previous answers: creates a y-length array every element of which is x, then takes the product. It's just gratuitously obfuscated to make it look like it's using the forbidden ** operator.

C, exponentiation by squaring

int power(int a, int b){
    if (b==0) return 1;
    if (b==1) return a;
    if (b%2==0) return power (a*a,b/2);
    return a*power(a*a,(b-1)/2);
}

golfed version in 46 bytes (thanks ugoren!)

p(a,b){return b<2?b?a:1:p(a*a,b/2)*(b&1?a:1);}

should be faster than all the other recursive answers so far o.O

slightly slower version in 45 bytes

p(a,b){return b<2?b?a:1:p(a*a,b/2)*p(a,b&1);}

Haskell - 55

pow x y=fix(\r a i->if i>=y then a else r(a*x)(i+1))1 0

There's already a shorter Haskell entry, but I thought it would be interesting to write one that takes advantage of the fix function, as defined in Data.Function. Used as follows (in the Repl for the sake of ease):

ghci> let pow x y=fix(\r a i->if i>=y then a else r(a*x)(i+1))1 0
ghci> pow 5 3
125

Q

9 chars. Generates array with y instances of x and takes the product.

{prd y#x}

Can explicitly cast to float for larger range given int/long x:

{prd y#9h$x}

Similar logic as many others, in PHP:

<?=array_product(array_fill(0,$argv[2],$argv[1]));

Run it with php file.php 5 3 to get 5^3

I'm not sure how many upvotes I can expect for this, but I found it somewhat peculiar that I actually had to write that very function today. And I'm pretty sure this is the first time any .SE site sees this language (website doesn't seem very helpful atm).

ABS

def Rat pow(Rat x, Int y) =
    if y < 0 then
        1 / pow(x, -y)
    else case y {
        0 => 1;
        _ => x * pow(x, y-1);
    };

Works for negative exponents and rational bases.

I highlighted it in Java syntax, because that's what I'm currently doing when I'm working with this language. Looks alright.

Pascal

The challenge did not specify the type or range of x and y, therefore I figure the following Pascal function follows all the given rules:

{ data type for a single bit: can only be 0 or 1 }
type
  bit = 0..1;

{ calculate the power of two bits, using the convention that 0^0 = 1 }
function bitpower(bit x, bit y): bit;
  begin
    if y = 0
      then bitpower := 1
      else bitpower := x
  end;

No loop, no built-in power or exponentiation function, not even recursion or arithmetics!

J - 5 or 4 bytes

Exactly the same as marinus' APL answer.

For x^y:

*/@$~

For y^x:

*/@$

For example:

   5 */@$~ 6
15625
   6 */@$ 5
15625

x $~ y creates a list of x repeated y times (same as y $ x

*/ x is the product function, */ 1 2 3 -> 1 * 2 * 3

Python

from math import sqrt

def pow(x, y):
    if y == 0:
        return 1
    elif y >= 1:
        return x * pow(x, y - 1)
    elif y > 0:
        y *= 2
        if y >= 1:
            return sqrt(x) * sqrt(pow(x, y % 1))
        else:
            return sqrt(pow(x, y % 1))
    else:
        return 1.0 / pow(x, -y)

Javascript

With tail recursion, works if y is a positive integer

function P(x,y,z){z=z||1;return y?P(x,y-1,x*z):z}

Bash

Everyone knows bash can do whizzy map-reduce type stuff ;-)

#!/bin/bash

x=$1
reduce () {
    ((a*=$x))
}
a=1
mapfile -n$2 -c1 -Creduce < <(yes)
echo $a

If thats too trolly for you then there's this:

#!/bin/bash

echo $(( $( yes $1 | head -n$2 | paste -s -d'*' ) ))

C

Yet another recursive exponentiation by squaring answer in C, but they do differ (this uses a shift instead of division, is slightly shorter and recurses one more time than the other):

e(x,y){return y?(y&1?x:1)*e(x*x,y>>1):1;}

Mathematica

This works for integers.

f[x_, y_] := Times@@Table[x, {y}]

Example

f[5,3]

125

How it works

Table makes a list of y x's. Times takes the product of all of them.`

Another way to achieve the same end:

#~Product~{i,1,#2}&

Example

#~Product~{i, 1, #2} & @@ {5, 3}

125

Windows Batch

Like most of the other answers here, it uses recursion.

@echo off
set y=%2
:p
if %y%==1 (
set z=%1
goto :eof
) else (
    set/a"y-=1"
    call :p %1
    set/a"z*=%1"
    goto :eof
)

x^y is stored in the environment variable z.

perl

Here's a tail recursive perl entry. Usage is echo $X,$Y | foo.pl:

($x,$y) = split/,/, <>;
sub a{$_*=$x;--$y?a():$_}
$_=1;
print a

Or for a more functional-type approach, how about:

($x,$y) = split/,/, <>;
$t=1; map { $t *= $x } (1..$y);
print $t

Python

def getRootOfY(x,y):
   return x**y 

def printAnswer():
   print "answer is ",getRootOfY(5,3)
printAnswer()

answer =125

I am not sure if this is against the requirements, but if not here is my attempt.