minix中atoi、atol、atof的实现

在minix2.0源代码中，有将字符串类型转换为int、long、double类型的函数实现，相关的实现函数分别在atoi.c、atol.c、atof.c文件中，我们来逐一学习其中的源码：

1、int atoi(register const char *nptr) ：将字符串类型转换为int类型

int atoi(register const char *nptr)
{int total = 0;int minus = 0;    //记录正负的变量（0：'+'，1：'-'）while (isspace(*nptr)) nptr++;  //滤去前导空格字符if (*nptr == '+') nptr++;else if (*nptr == '-') {minus = 1;nptr++;}while (isdigit(*nptr)) {total *= 10;total += (*nptr++ - '0');}return minus ? -total : total;
}

2、long atol(register const char *nptr) ：将字符串类型转换为long类型，与atoi极为类似

 1 long atol(register const char *nptr)
 2 {
 3     long total = 0;
 4     int minus = 0;
 5 
 6     while (isspace(*nptr)) nptr++;
 7     if (*nptr == '+') nptr++;
 8     else if (*nptr == '-') {
 9         minus = 1;
10         nptr++;
11     }
12     while (isdigit(*nptr)) {
13         total *= 10;
14         total += (*nptr++ - '0');
15     }
16     return minus ? -total : total;
17 }

3、double atof(const char *nptr)：将字符串类型转换为double类型

 1 double atof(const char *nptr)
 2 {
 3     double d;
 4     int e = errno;
 5 
 6     d = strtod(nptr, (char **) NULL); 
 7 
 8      //strtod:  stdlib.h中定义的库函数，将字符串转换为double型
 9     //double strtod(const char * restrict nptr, char ** restrict endptr); 
10     errno = e;
11     return d;
12 }

PS：在《C程序设计语言》中Ritchie提供了一种atof的实现：

 1 double atof(char *s)
 2 {
 3     double val, power;
 4     int sign, i;
 5 
 6     for (i = 0; isspace(s[i]); i++)
 7         ;
 8     sign = (s[i] == '-') ? -1 : 1;
 9     if (s[i] == '+' || s[i] == '-')
10         i++;
11     for (val = 0.0; isdigit(s[i]); i++)
12         val = 10.0 * val + (s[i] - '0');
13     if (s[i] == '.')
14         i++;
15     for (power = 1.0; isdigit(s[i]); i++) {
16         val = 10.0 * val + (s[i] - '0');
17         power *= 10.0;
18     }
19     return sign * val / power;
20 }

后面的练习4-2，要求我们对atof函数进行扩展，使它能够处理形如123.45e-6这样的科学表示法

 1 #include<stdio.h>
 2 
 3 double atof(char *s)
 4 {
 5     double val, power, temp, exp;
 6     char flag;
 7     int sign, i;
 8 
 9     for (i = 0; isspace(s[i]); i++)
10         ;
11     sign = (s[i] == '-') ? -1 : 1;
12     if (s[i] == '+' || s[i] == '-')
13         i++;
14     for (val = 0.0; isdigit(s[i]); i++)
15         val = 10.0 * val + (s[i] - '0');
16     if (s[i] == '.')
17         i++;
18     for (power = 1.0; isdigit(s[i]); i++) {
19         val = 10.0 * val + (s[i] - '0');
20         power *= 10.0;
21     }
22     temp = sign * val / power;
23     if (s[i] == 'e' || s[i] == 'E')
24         i++;
25     flag = s[i];
26     if (flag == '-' || flag == '+')
27         i++;
28     for (exp = 0; isdigit(s[i]) && s[i] !='\0'; i++) {
29         exp = 10.0 * exp + (s[i] - '0');
30     }
31     if (flag == '+') {
32         while(exp--) temp *= 10;
33     } else {
34         while(exp--) temp /= 10;
35     }
36     return temp;
37 }
38 
39 int main()
40 {
41     char a[] = "  -2309.12E-15";
42     printf("%e\n",atof(a));
43     return 0;
44 
45 }