通过位操作实现四则运算
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
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