atoi和itoa函数实现

转：http://www.wypblog.com/archives/224

atoi函数是C语言库提供的，是把字符串转换成整型数和把字符串转换成整型数。而itoa函数是广泛应用的非标准C语言扩展函数,由于它不是标准C语言函数，所以不能在所有的编译器中使用，它的功能是把一整数转换为字符串。
两个函数功能很好理解，但是它们的实现需要考虑到很多问题，在面试中，很多面试官都会问atoi和itoa的实现，这可以很好的了解程序员编程的功底。
那么在实现atoi一般需要考虑到那些情况呢？

1. 首先是整数，很多人很自然的会忽略负整数。
2. 其次是考虑到上溢和下溢
3. 字符串以空格开始怎么去处理
4. 字符串中包含了除0-9之间的字符该怎么去处理

实现

#include <stdio.h>#define MAX_INT ((1 << 31) - 1)#define MIN_INT (-(1 << 31))int atoi(const char *str){char *temp = str;int i = 0;int flags = 0;unsigned int sum = 0;while(*temp == ' ') ++temp ;if(*temp != '-' && *temp != '+' && (*temp < '0' || *temp > '9')){//第一个字符不是数字return 0;}if(*temp == '-'){ //第一个是负号flags = 1;++temp;}else if(*temp == '+'){++temp;}while(*temp >= '0' && *temp <= '9'){if(!flags){//上溢if(sum > MAX_INT / 10 || (sum == MAX_INT / 10 && (*temp > '7'))){return MAX_INT;}}else{//下溢if(sum > MAX_INT / 10 || (sum == MAX_INT / 10 && (*temp > '8'))){return MIN_INT;}}sum = sum * 10 + (*temp - '0');++temp;}//if(flags){//sum *= -1;//}return flags ? (-1 * sum) : sum;}int main(){printf("%d\n", atoi("    0125464 c 646"));return 0;}

ansi库中的实现：

#include<ctype.h>#include<stdlib.h>intatoi(register const char *nptr){return strtol(nptr, (char **) NULL, 10);}/* * (c) copyright 1987 by the Vrije Universiteit, Amsterdam, The Netherlands. * See the copyright notice in the ACK home directory, in the file "Copyright". *//* $Header: strtol.c,v 1.4 90/05/11 15:22:19 eck Exp $ */#include<ctype.h>#include<errno.h>#include<limits.h>#include<stdlib.h>static unsigned longstring2long(register const char *nptr, char **endptr,int base, int is_signed);long intstrtol(register const char *nptr, char **endptr, int base){return (signed long)string2long(nptr, endptr, base, 1);}#define between(a, c, z)  ((unsigned) ((c) - (a)) <= (unsigned) ((z) - (a)))static unsigned longstring2long(register const char *nptr, char ** const endptr,int base, int is_signed){register unsigned int v;register unsigned long val = 0;register int c;int ovfl = 0, sign = 1;const char *startnptr = nptr, *nrstart;if (endptr) *endptr = (char *)nptr;while (isspace(*nptr)) nptr++;c = *nptr;if (c == '-' || c == '+') {if (c == '-') sign = -1;nptr++;}nrstart = nptr;/* start of the number *//* When base is 0, the syntax determines the actual base */if (base == 0)if (*nptr == '0')if (*++nptr == 'x' || *nptr == 'X') {base = 16;nptr++;}elsebase = 8;elsebase = 10;else if (base==16 && *nptr=='0' && (*++nptr =='x' || *nptr =='X'))nptr++;for (;;) {c = *nptr;if (between('0', c, '9')) {v = c - '0';} elseif (between('a', c, 'z')) {v = c - 'a' + 0xa;} elseif (between('A', c, 'Z')) {v = c - 'A' + 0xA;} else {break;}if (v >= base) break;if (val > (ULONG_MAX - v) / base) ovfl++;val = (val * base) + v;nptr++;}if (endptr) {if (nrstart == nptr) *endptr = (char *)startnptr;else *endptr = (char *)nptr;}if (!ovfl) {/* Overflow is only possible when converting a signed long. */if (is_signed    && (   (sign < 0 && val > -(unsigned long)LONG_MIN)|| (sign > 0 && val > LONG_MAX)))    ovfl++;}if (ovfl) {errno = ERANGE;if (is_signed)if (sign < 0) return LONG_MIN;else return LONG_MAX;else return ULONG_MAX;}return (long) sign * val;}

itoa实现注意事项：

1. 忘记考虑负数；
2. 忘记在末尾加上’\0′。

实现：

#include <stdlib.h>　　#include <stdio.h> char *itoa(int num,char *str,int radix){//num：int型原数,str:需转换成的string，radix,原进制， /* 索引表 */char index[]="0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";unsigned unum;/* 中间变量 */int i=0,j,k;/* 确定unum的值 */if(radix==10&&num<0){/* 十进制负数 */unum=(unsigned)-num;str[i++]='-';}elseunum=(unsigned)num;/* 其他情况 *//* 逆序 */do{str[i++]=index[unum%(unsigned)radix];unum/=radix;}while(unum);str[i]='\0';/* 转换 */if(str[0]=='-')k=1;/* 十进制负数 */elsek=0;/* 将原来的“/2”改为“/2.0”，保证当num在16~255之间，radix等于16时，也能得到正确结果 */for(j=k;j<(i-1)/2.0+k;j++){num=str[j];str[j]=str[i-j-1+k];str[i-j-1+k]=num;}return str;}int main(){int number = 01777777777777777777774; char string[25]; itoa(number, string, 10); printf("integer = %d string = %s\n", number, string); return 0; }

ansi库实现

#define NUMBER_OF_DIGITS 16void _uitoa(unsigned int value, char* string, unsigned char radix){unsigned char index, i;/* char buffer[NUMBER_OF_DIGITS]; */ /* space for NUMBER_OF_DIGITS + '\0' */  index = NUMBER_OF_DIGITS;  i = 0;  do {    string[--index] = '0' + (value % radix);    if ( string[index] > '9') string[index] += 'A' - '9' - 1;    value /= radix;  } while (value != 0);  do {    string[i++] = string[index++];  } while ( index < NUMBER_OF_DIGITS );  string[i] = 0; /* string terminator */}void _itoa(int value, char* string, unsigned char radix){  if (value < 0 && radix == 10) {    *string++ = '-';    _uitoa(-value, string, radix);  }  else {    _uitoa(value, string, radix);  }}