C說話完成年夜整數加減運算詳解。本站提示廣大學習愛好者:(C說話完成年夜整數加減運算詳解)文章只能為提供參考,不一定能成為您想要的結果。以下是C說話完成年夜整數加減運算詳解正文
媒介
我們曉得,在數學中,數值的年夜小是沒有下限的,然則在盤算機中,因為字長的限制,盤算機所能表現的規模是無限的,當我們比較較小的數停止運算時,如:1234+5678,如許的數值並沒有超越盤算機的表現規模,所以可以運算。然則當我們在現實的運用中停止年夜量的數據處置時,會發明介入運算的數常常跨越盤算機的根本數據類型的表現規模,好比說,在地理學上,假如一個星球間隔我們為100萬光年,那末我們將其化簡為千米,或許是米的時刻,我們會發明這是一個很年夜的數。如許盤算機將沒法對其停止直接盤算。
能夠我們以為現實運用中的年夜數也不外就是幾百位罷了,現實上,在某些范疇裡,乃至能夠湧現幾百萬位的數據停止運算,這是我們很難想象的。假如沒有盤算機,那末盤算效力可想而知。
因為編程說話供給的根本數值數據類型表現的數值規模無限,不克不及知足較年夜范圍的高精度數值盤算,是以須要應用其他辦法完成高精度數值的盤算,因而發生了年夜數運算。本項目完成了年夜數運算的加、減運算。
一. 成績提出
用C說話完成一個年夜整數盤算器。初步請求支撐年夜整數的加、減運算,例如8888888888888+1112=8888888890000或1000000000000-999999999999=1。
C說話中,整型變量所能存儲的最寬數據為0xFFFF FFFF,對應的無符號數為4294967295,即沒法保留跨越10位的整數。留意,此處"10位"指數學中的10個數字,並不是盤算機迷信中的10比特。浮點類型double固然可以存儲更多位數的整數,但一方面常數字面量寬度受編譯器限制,另外一方面經由過程浮點方法處置整數精度較低。例如:
double a = 1377083362513770833626.0, b=1585054852315850548524.0; printf("res = %.0f\n", a+b);
輸入為res = 2962138214829621510144
,而准確值應為2962138214829621382150。
既然根本數據類型沒法表現年夜整數,那末只能本身設計存儲方法來完成年夜整數的表現和運算。平日,輸出的年夜整數為字符串情勢。是以,罕見的思緒是將年夜整數字符串轉化為數組,再用數組模仿年夜整數的運算。詳細而言,先將字符串中的數字字符次序存入一個較年夜的整型數組,其元素代表整數的某一名或某幾位(如萬進制);然後依據運算規矩操作數組元素,以模仿整數運算;最初,將數組元素次序輸入。
數組方法操作便利,完成簡略,缺陷是空間應用率和履行效力不高。也可直接操作年夜整數字符串,從字符串末尾逆向盤算。本文完成就采取這類方法。
二. 代碼完成
起首,給出幾個宏界說和運算構造:
#include<stdio.h> #include<stdlib.h> #include<string.h> #define ADD_THRES (sizeof("4294967295")-2) //兩個9位整數相加不會溢出 #define MUL_THRES (sizeof("65535")-2) //兩個4位整數相乘不會溢出 #define OTH_THRES (sizeof("4294967295")-1) //兩個10位整數相減或相除不會溢出 typedef struct{ char *leftVal; char *rightVal; char operator; }MATH_OPER;
基於上述界說,以下將順次給出運算代碼的完成。
加法運算重要存眷相加進程中的進位成績:
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 } }
留意第33~36行的處置,當最高位未定期望發生進位時,本來為0的resBuf[0]被置為空格字符,不然將沒法輸入運算成果。固然,也可將resBuf全體前移一個元素。
減法運算絕對龐雜,須要依據被減數和減數的年夜小調劑運算次序。若被減數小於減數("11-111"或"110-111"),則交流被減數和減數後再做正常的減法運算,而且成果需添加負號前綴。另外,還需存眷借位成績。
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'; }
關於Addition()
和Subtraction()
函數,設計測試用例以下:
#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"); }
斟酌到說話內置的運算效力應當更高,是以在弗成能發生溢出時盡可能選用內置運算。CalcOperation()
函數便采取這一思緒:
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; }
留意,年夜整數的乘法和除法運算還沒有完成,是以響應代碼分支直接前往。
最初,完成進口函數:
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; }
上述代碼中,scanf()
函數的格局化字符串作風相似正則表達式。其具體引見拜見《sscanf的字符串格局化用法》一文。
三. 後果驗證
將上節代碼存為BigIntOper.c
文件。測試成果以下:
[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 ~]$
經由過程外部測試用例和內部人工校驗,可知運算成果准確無誤。
總結
以上就是C說話完成年夜整數加減運算的全體內容,年夜家都學會了嗎?願望本文的內容對年夜家進修C說話能有所贊助。