用基本算術運算實現指數函數

Question

爲了練習的目的，我必須用最基本的算術運算來實現指數函數。我想出了這個，其中X是基礎和ÿ指數：

function expAetB() { 
    product=1; 
    for (i=0; i<y; i++) 
    { 
      product=product*x; 
    } 
    return product; 
}; 

然而，還有比product=product*x;更基本的操作。我應該以某種方式能夠插入，而不是另一個for循環乘法和傳遞的結果，但我不能找到一種方式來做到這一點，而不陷入無限循環。

Answer 1

以相同的方式，指數是重複的乘法，所以乘法就是簡單的重複加法。

簡單地創建另一個函數mulAetB它爲你做，並注意諸如負面投入的東西。

你甚至可以進一步級別並定義增量和減量，但這可能是矯枉過正。

---

見，例如，其使用另外的矯枉過正方法下面的程序：

#include <stdio.h> 

static unsigned int add (unsigned int a, unsigned int b) { 
    unsigned int result = a; 
    while (b-- != 0) result++; 
    return result; 
} 

static unsigned int mul (unsigned int a, unsigned int b) { 
    unsigned int result = 0; 
    while (b-- != 0) result = add (result, a); 
    return result; 
} 

static unsigned int pwr (unsigned int a, unsigned int b) { 
    unsigned int result = 1; 
    while (b-- != 0) result = mul (result, a); 
    return result; 
} 

int main (void) { 
    int test[] = {0,5, 1,9, 2,4, 3,5, 7,2, -1}, *ip = test; 
    while (*ip != -1) { 
     printf ("%d + %d = %3d\n" , *ip, *(ip+1), add (*ip, *(ip+1))); 
     printf ("%d x %d = %3d\n" , *ip, *(ip+1), mul (*ip, *(ip+1))); 
     printf ("%d^%d = %3d\n\n", *ip, *(ip+1), pwr (*ip, *(ip+1))); 
     ip += 2; 
    } 
    return 0; 
} 

該程序的輸出顯示計算是正確的：

0 + 5 = 5 
0 x 5 = 0 
0^5 = 0 

1 + 9 = 10 
1 x 9 = 9 
1^9 = 1 

2 + 4 = 6 
2 x 4 = 8 
2^4 = 16 

3 + 5 = 8 
3 x 5 = 15 
3^5 = 243 

7 + 2 = 9 
7 x 2 = 14 
7^2 = 49 

---

如果你真的必須它在一個單一的功能，這是一個簡單的問題重構函數調用是內聯：

static unsigned int pwr (unsigned int a, unsigned int b) { 
    unsigned int xres, xa, result = 1; 

    // Catch common cases, simplifies rest of function (a>1, b>0) 

    if (b == 0) return 1; 
    if (a == 0) return 0; 
    if (a == 1) return 1; 

    // Do power as repeated multiplication. 

    result = a; 
    while (--b != 0) { 
     // Do multiplication as repeated addition. 

     xres = result; 
     xa = a; 
     while (--xa != 0) 
      result = result + xres; 
    } 

    return result; 
}