大整数加减运算的C语言实现

大整数加减运算的C语言实现

标签: 大整数加减 C


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=88888888900001000000000000-999999999999=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。

posted @ 2016-06-22 15:37  clover_toeic  阅读(8461)  评论(0编辑  收藏