在最早学习四则运算的过程中，我们其实就已经掌握了进制算法，这一次我将对二进制运用这个进制算法来实现四则运算。
四则运算
math.c
/**
* 功能：通过位操作实现四则运算
* 算法：完全参照十进制的进制算法
*
* Created with CLion
* User: zzzz76
* Date: 2018-02-10
*/
#include <stdio.h>
#include <assert.h>
#include "math.h"
/**
* 加法：往上递归实现
*
* @param a
* @param b
* @return
*/
int base_add(int a, int b) {
/* 值在函数中的传递只能通过参数或者返回操作，所以递归效果无非是体现在参数和返回操作上 */
if (b == 0) {
return a;
}
int save = a ^b;
int promote = (a & b) << 1;
return base_add(save, promote);
}
/**
* 加法：迭代实现
*
* @param a
* @param b
* @return
*/
int base_add_re(int a, int b) {
while (a && b) {
int promote = (a & b) << 1;
a = a ^ b;
b = promote;
}
return a ^ b;
}
/**
* 减法：往上递归实现
*
* @param a
* @param b
* @return
*/
int base_sub(int a, int b) {
if (b == 0) {
return a;
}
int save = a ^ b;
int reduce = ((~a) & b) << 1;
return base_sub(save, reduce);
}
/**
* 减法：迭代实现
*
* @param a
* @param b
* @return
*/
int base_sub_re(int a, int b) {
while(b) {
int save = a ^ b;
b = ((~a) & b) << 1;
a = save;
}
return a;
}
/**
* 减法：补位实现
*
* @param a
* @param b
* @return
*/
int base_sub_re_re(int a, int b) {
return base_add(a, base_add(~b, 1));
}
/**
* 乘法：递归实现
*
* @param a
* @param b
* @return
*/
int base_mul(int a, int b) {
int count = 0;
if (a == 0) {
return count;
}
if (a & 1) {
count = base_add(count, b);
}
a = (unsigned)a >> 1;
b <<= 1;
count = base_add(count, base_mul(a, b));
return count;
}
/**
* 乘法：迭代实现
*
* @param a
* @param b
* @return
*/
int base_mul_re(int a, int b) {
int count = 0;
while (a) {
if (a & 1) {
count = base_add(count, b);
}
a = (unsigned)a >> 1;
b <<= 1;
}
return count;
}
/**
* 除法：迭代实现
*
* @param a
* @param b
* @return
*/
int base_div(int a, int b) {
assert(b);
int result = 0;
int bit_num = 31;
while (bit_num != -1) {
if (b <= ((unsigned) a >> bit_num)) {
result = base_add(result, 1 << bit_num);
a = base_sub(a, b << bit_num);
}
bit_num = base_sub(bit_num, 1);
}
return result;
}
/**
* 除法：递归实现
*
* @param a
* @param b
* @return
*/
int base_div_re(int a, int b) {
assert(b);
if (a < (unsigned) b) {
return 0;
}
int bit_num = 0;
while (b <= (unsigned) a >> 1 >> bit_num) {
bit_num = base_add(bit_num, 1);
}
int result = 1 << bit_num;
a = base_sub(a, b << bit_num);
result = base_add(result, base_div_re(a, b));
return result;
}
math.h
/**
* Created with CLion
* User: zzzz76
* Date: 2018-02-12
*/
#ifndef MATH_H
#define MATH_H
int base_add(int a, int b);
int base_add_re(int a, int b);
int base_sub(int a, int b);
int base_sub_re(int a, int b);
int base_sub_re_re(int a, int b);
int base_mul(int a, int b);
int base_mul_re(int a, int b);
int base_div(int a, int b);
int base_div_re(int a, int b);
#endif //MATH_H
test.c
/**
* Created with CLion
* User: zzzz76
* Date: 2018-02-12
*/
#include "math.h"
#include <stdio.h>
static int main_ret = 0;
static int test_count = 0;
static int test_pass = 0;
#define EXPECT_EQ_BASE(expect, actual, format) do { test_count++; if ((expect) == (actual)) { test_pass++; } else { fprintf(stderr, "%s:%d: expect: " format " actual: " format "\n", __FILE__, __LINE__, expect, actual); main_ret = 1; } } while(0);
#define EXPECT_EQ_INT(expect, actual) EXPECT_EQ_BASE(expect, actual, "%d");
#define TEST_BASE_ADD(expect, a, b) EXPECT_EQ_INT(expect, base_add(a, b)); EXPECT_EQ_INT(expect, base_add_re(a, b));
static void test_base_add() {
TEST_BASE_ADD(2, 1, 1);
TEST_BASE_ADD(0, -1, 1);
TEST_BASE_ADD(0, 1, -1);
TEST_BASE_ADD(-2, -1, -1);
TEST_BASE_ADD(-2147483648, 2147483647, 1);
}
#define TEST_BASE_SUB(expect, a, b) EXPECT_EQ_INT(expect, base_sub(a, b)); EXPECT_EQ_INT(expect, base_sub_re(a, b)); EXPECT_EQ_INT(expect, base_sub_re_re(a, b));
static void test_base_sub() {
TEST_BASE_SUB(0, 1, 1);
TEST_BASE_SUB(-2, -1, 1);
TEST_BASE_SUB(2, 1, -1);
TEST_BASE_SUB(0, -1, -1);
TEST_BASE_SUB(2147483647, -2147483648, 1);
}
#define TEST_BASE_MUL(expect, a, b) EXPECT_EQ_INT(expect, base_mul(a, b)); EXPECT_EQ_INT(expect, base_mul_re(a, b));
static void test_base_mul() {
TEST_BASE_MUL(9, 3, 3);
TEST_BASE_MUL(-9, -3, 3);
TEST_BASE_MUL(-9, 3, -3);
TEST_BASE_MUL(9, -3, -3);
TEST_BASE_MUL(0, -2147483648, 2);
TEST_BASE_MUL(-2, 2147483647, 2);
}
#define TEST_BASE_DIV(expect, a, b) EXPECT_EQ_INT(expect, base_div(a, b)); EXPECT_EQ_INT(expect, base_div_re(a, b));
static void test_base_div() {
TEST_BASE_DIV(2, 2, 1);
TEST_BASE_DIV(-2, -2, 1);
TEST_BASE_DIV(0, 2, -1);
TEST_BASE_DIV(0, -2, -1);
TEST_BASE_DIV(0, 1, 2);
TEST_BASE_DIV(0, 1, -2);
TEST_BASE_DIV(2147483647, -1, 2);
TEST_BASE_DIV(1, -1, -2);
}
static void test_base() {
test_base_add();
test_base_sub();
test_base_mul();
test_base_div();
}
int main() {
test_base();
printf("%d/%d (%3.2f%%) passed\n", test_pass, test_count, test_pass * 100.0 / test_count);
return main_ret;
}
从递归角度看待代码
递归是函数调用，考虑值的传递过程，适合阅读。
迭代是具体的实现过程，往往代码效率更加充分。
通常我们在写代码时，往往注重代码的效率和正确性，而忽略了代码表达的意思，致使代码难以阅读。所以对两者的取舍，一定程度影响了代码的好与坏