大整数加减运算的C语言实现

标签：大整数加减 C

大整数加减运算的C语言实现
一. 问题提出
二. 代码实现
三. 效果验证

一. 问题提出

培训老师给出一个题目：用C语言实现一个大整数计算器。初步要求支持大整数的加、减运算，例如8888888888888+1112=8888888890000或1000000000000-999999999999=1。

C语言中，整型变量所能存储的最宽数据为0xFFFF FFFF，对应的无符号数为4294967295，即无法保存超过10位的整数。注意，此处"10位"指数学中的10个数字，并非计算机科学中的10比特。浮点类型double虽然可以存储更多位数的整数，但一方面常数字面量宽度受编译器限制，另一方面通过浮点方式处理整数精度较低。例如：

    double a = 1377083362513770833626.0, b=1585054852315850548524.0;

    printf("res = %.0f\n", a+b);

输出为res = 2962138214829621510144，而正确值应为2962138214829621382150。

既然基本数据类型无法表示大整数，那么只能自己设计存储方式来实现大整数的表示和运算。通常，输入的大整数为字符串形式。因此，常见的思路是将大整数字符串转化为数组，再用数组模拟大整数的运算。具体而言，先将字符串中的数字字符顺序存入一个较大的整型数组，其元素代表整数的某一位或某几位(如万进制)；然后根据运算规则操作数组元素，以模拟整数运算；最后，将数组元素顺序输出。

数组方式操作方便，实现简单，缺点是空间利用率和执行效率不高。也可直接操作大整数字符串，从字符串末尾逆向计算。本文实现就采用这种方式。

二. 代码实现

首先，给出几个宏定义和运算结构：

#include<stdio.h>

#include<stdlib.h>

#include<string.h>

#define ADD_THRES     (sizeof("4294967295")-2)  //两个9位整数相加不会溢出

#define MUL_THRES     (sizeof("65535")-2)       //两个4位整数相乘不会溢出

#define OTH_THRES     (sizeof("4294967295")-1)  //两个10位整数相减或相除不会溢出

typedef struct{

    char *leftVal;

    char *rightVal;

    char operator;

}MATH_OPER;

基于上述定义，以下将依次给出运算代码的实现。

加法运算主要关注相加过程中的进位问题：

void Addition(char *leftVal,  char *rightVal,

              char *resBuf, unsigned int resbufLen) {

    unsigned int leftLen = strlen(leftVal);

    unsigned int rightLen = strlen(rightVal);

    unsigned char isLeftLonger = (leftLen>=rightLen) ? 1 : 0;

    unsigned int longLen = isLeftLonger ? leftLen : rightLen;

    if(resbufLen < longLen) { //possible carry + string terminator

        fprintf(stderr, "Not enough space for result(cur:%u)!\n", resbufLen);

        return;

    }

    char *longAddend = isLeftLonger ? leftVal : rightVal;

    char *shortAddend = isLeftLonger ? rightVal : leftVal;

    unsigned int diffLen = isLeftLonger ? (leftLen-rightLen) : (rightLen-leftLen);

    //a carry might be generated from adding the most significant digit

    if((leftLen == rightLen) && (leftVal[0]-'0'+rightVal[0]-'0' >= 9))

        resBuf += 1;

    unsigned int carry = 0;

    int i = longLen-1;

    for(; i >= 0; i--) {

        unsigned int leftAddend = longAddend[i] - '0';

        unsigned int rightAddend = (i<diffLen) ? 0 : shortAddend[i-diffLen]-'0';

        unsigned int digitSum = leftAddend + rightAddend + carry;

        resBuf[i] = digitSum % 10 + '0';

        carry = (digitSum >= 10) ? 1 : 0;

    }

    if(carry == 1) {

        resBuf -= 1;

        resBuf[0] = '1';

    }

    else if(leftVal[0]-'0'+rightVal[0]-'0' == 9) {

        resBuf -= 1;

        resBuf[0] = ' '; //fail to generate a carry

    }

}

注意第33~36行的处理，当最高位未按期望产生进位时，原来为0的resBuf[0]被置为空格字符，否则将无法输出运算结果。当然，也可将resBuf整体前移一个元素。

减法运算相对复杂，需要根据被减数和减数的大小调整运算顺序。若被减数小于减数("11-111"或"110-111")，则交换被减数和减数后再做正常的减法运算，并且结果需添加负号前缀。此外，还需关注借位问题。

void Subtraction(char *leftVal,  char *rightVal,

                 char *resBuf, unsigned int resbufLen) {

    int cmpVal = strcmp(leftVal, rightVal);

    if(!cmpVal) {

        resBuf[0] = '0';

        return;

    }

    unsigned int leftLen = strlen(leftVal);

    unsigned int rightLen = strlen(rightVal);

    unsigned char isLeftLonger = 0;

    if((leftLen > rightLen) ||              //100-10

       (leftLen == rightLen && cmpVal > 0)) //100-101

        isLeftLonger = 1;

    unsigned int longLen = isLeftLonger ? leftLen : rightLen;

    if(resbufLen <= longLen) { //string terminator

        fprintf(stderr, "Not enough space for result(cur:%u)!\n", resbufLen);

        return;

    }

    char *minuend = isLeftLonger ? leftVal : rightVal;

    char *subtrahend = isLeftLonger ? rightVal : leftVal;

    unsigned int diffLen = isLeftLonger ? (leftLen-rightLen) : (rightLen-leftLen);

    //a borrow will be generated from subtracting the most significant digit

    if(!isLeftLonger) {

        resBuf[0] = '-';

        resBuf += 1;

    }

    unsigned int borrow = 0;

    int i = longLen-1;

    for(; i >= 0; i--)

    {

        unsigned int expanSubtrahend = (i<diffLen) ? '0' : subtrahend[i-diffLen];

        int digitDif = minuend[i] - expanSubtrahend - borrow;

        borrow = (digitDif < 0) ? 1 : 0;

        resBuf[i] = digitDif + borrow*10 + '0';

        //printf("[%d]Dif=%d=%c-%c-%d -> %c\n", i, digitDif, minuend[i], expanSubtrahend, borrow, resBuf[i]);

    }

    //strip leading '0' characters

    int iSrc = 0, iDst = 0, isStripped = 0;

    while(resBuf[iSrc] !='\0') {

        if(isStripped) {

            resBuf[iDst] = resBuf[iSrc];

            iSrc++; iDst++;

        }

        else if(resBuf[iSrc] != '0') {

            resBuf[iDst] = resBuf[iSrc];

            iSrc++; iDst++;

            isStripped = 1;

        }

        else

            iSrc++;

     }

     resBuf[iDst] = '\0';

}

对于Addition()和Subtraction()函数，设计测试用例如下：

#include<assert.h>

#define ASSERT_ADD(_add1, _add2, _sum) do{\

    char resBuf[100] = {0}; \

    Addition(_add1, _add2, resBuf, sizeof(resBuf)); \

    assert(!strcmp(resBuf, _sum)); \

}while(0)

#define ASSERT_SUB(_minu, _subt, _dif) do{\

    char resBuf[100] = {0}; \

    Subtraction(_minu, _subt, resBuf, sizeof(resBuf)); \

    assert(!strcmp(resBuf, _dif)); \

}while(0)

void VerifyOperation(void) {

    ASSERT_ADD("22", "1686486458", "1686486480");

    ASSERT_ADD("8888888888888", "1112", "8888888890000");

    ASSERT_ADD("1234567890123", "1", "1234567890124");

    ASSERT_ADD("1234567890123", "3333333333333", "4567901223456");

    ASSERT_ADD("1234567890123", "9000000000000", "10234567890123");

    ASSERT_ADD("1234567890123", "8867901223000", "10102469113123");

    ASSERT_ADD("1234567890123", "8000000000000", " 9234567890123");

    ASSERT_ADD("1377083362513770833626", "1585054852315850548524", "2962138214829621382150");

    ASSERT_SUB("10012345678890", "1", "10012345678889");

    ASSERT_SUB("1", "10012345678890", "-10012345678889");

    ASSERT_SUB("10012345678890", "10012345678891", "-1");

    ASSERT_SUB("10012345678890", "10012345686945", "-8055");

    ASSERT_SUB("1000000000000", "999999999999", "1");

}

考虑到语言内置的运算效率应该更高，因此在不可能产生溢出时尽量选用内置运算。CalcOperation()函数便采用这一思路：

void CalcOperation(MATH_OPER *mathOper, char *resBuf, unsigned int resbufLen) {

    unsigned int leftLen = strlen(mathOper->leftVal);

    unsigned int rightLen = strlen(mathOper->rightVal);

    switch(mathOper->operator) {

        case '+':

            if(leftLen <= ADD_THRES && rightLen <= ADD_THRES)

                snprintf(resBuf, resbufLen, "%d",

                         atoi(mathOper->leftVal) + atoi(mathOper->rightVal));

            else

                Addition(mathOper->leftVal, mathOper->rightVal, resBuf, resbufLen);

            break;

        case '-':

            if(leftLen <= OTH_THRES && rightLen <= OTH_THRES)

                snprintf(resBuf, resbufLen, "%d",

                         atoi(mathOper->leftVal) - atoi(mathOper->rightVal));

            else

                Subtraction(mathOper->leftVal, mathOper->rightVal, resBuf, resbufLen);

            break;

        case '*':

            if(leftLen <= MUL_THRES && rightLen <= MUL_THRES)

                snprintf(resBuf, resbufLen, "%d",

                         atoi(mathOper->leftVal) * atoi(mathOper->rightVal));

            else

                break; //Multiplication: product = multiplier * multiplicand

            break;

        case '/':

            if(leftLen <= OTH_THRES && rightLen <= OTH_THRES)

                snprintf(resBuf, resbufLen, "%d",

                         atoi(mathOper->leftVal) / atoi(mathOper->rightVal));

            else

                break; //Division: quotient = dividend / divisor

            break;

        default:

            break;

    }

    return;

}

注意，大整数的乘法和除法运算尚未实现，因此相应代码分支直接返回。

最后，完成入口函数：

int main(void) {

    VerifyOperation();

    char leftVal[100] = {0}, rightVal[100] = {0}, operator='+';

    char resBuf[1000] = {0};

    //As you see, basically any key can quit:)

    printf("Enter math expression(press q to quit): ");

    while(scanf(" %[0-9] %[+-*/] %[0-9]", leftVal, &operator, rightVal) == 3) {

        MATH_OPER mathOper = {leftVal, rightVal, operator};

        memset(resBuf, 0, sizeof(resBuf));

        CalcOperation(&mathOper, resBuf, sizeof(resBuf));

        printf("%s %c %s = %s\n", leftVal, operator, rightVal, resBuf);

        printf("Enter math expression(press q to quit): ");

    }

    return 0;

}

上述代码中，scanf()函数的格式化字符串风格类似正则表达式。其详细介绍参见《sscanf的字符串格式化用法》一文。

三. 效果验证

将上节代码存为BigIntOper.c文件。测试结果如下：

[wangxiaoyuan_@localhost ~]$ gcc -Wall -o BigIntOper BigIntOper.c

[wangxiaoyuan_@localhost ~]$ ./BigIntOper

Enter math expression(press q to quit): 100+901

100 + 901 = 1001

Enter math expression(press q to quit): 100-9

100 - 9 = 91

Enter math expression(press q to quit): 1234567890123 + 8867901223000

1234567890123 + 8867901223000 = 10102469113123

Enter math expression(press q to quit): 1377083362513770833626 - 1585054852315850548524

1377083362513770833626 - 1585054852315850548524 = -207971489802079714898

Enter math expression(press q to quit): q

[wangxiaoyuan_@localhost ~]$

通过内部测试用例和外部人工校验，可知运算结果正确无误。