面试题strtoi实现（一）—— 函数的简单实现

大师级程序地址： http://www.opensource.apple.com/source/xnu/xnu-1456.1.26/bsd/libkern/strtol.c

[cpp] view plaincopyprint?

1. static inlineint 
2. isupper(char c) 
3. { 
4.     return (c >= 'A' && c <='Z'); 
5. } 
6.  
7. static inlineint 
8. isalpha(char c) 
9. { 
10.     return ((c >= 'A' && c <='Z') || (c >= 'a' && c <='z')); 
11. } 
12.  
13.  
14. static inlineint 
15. isspace(char c) 
16. { 
17.     return (c == ' ' || c == '\t' || c =='\n' || c == '\12'); 
18. } 
19.  
20. static inlineint 
21. isdigit(char c) 
22. { 
23.     return (c >= '0' && c <= '9'); 
24. } 
25.  
26. /*
27. * Convert a string to a long integer.
28. *
29. * Ignores `locale' stuff.  Assumes that the upper and lower case
30. * alphabets and digits are each contiguous.
31. */ 
32. long 
33. strtol(nptr, endptr, base) 
34.     const char *nptr; 
35.     char **endptr; 
36.     register int base; 
37. { 
38.     register constchar *s = nptr; 
39.     register unsigned long acc; 
40.     register int c; 
41.     register unsigned long cutoff; 
42.     register int neg = 0, any, cutlim; 
43.  
44.     /*
45.      * Skip white space and pick up leading +/- sign if any.
46.      * If base is 0, allow 0x for hex and 0 for octal, else
47.      * assume decimal; if base is already 16, allow 0x.
48.      */ 
49.     do { 
50.         c = *s++; 
51.     } while (isspace(c)); 
52.     if (c == '-') { 
53.         neg = 1; 
54.         c = *s++; 
55.     } else if (c =='+') 
56.         c = *s++; 
57.     if ((base == 0 || base == 16) && 
58.         c == '0' && (*s == 'x' || *s =='X')) { 
59.         c = s[1]; 
60.         s += 2; 
61.         base = 16; 
62.     } else if ((base == 0 || base == 2) && 
63.         c == '0' && (*s =='b' || *s == 'B')) { 
64.         c = s[1]; 
65.         s += 2; 
66.         base = 2; 
67.     } 
68.     if (base == 0) 
69.         base = c == '0' ? 8 : 10; 
70.  
71.     /*
72.      * Compute the cutoff value between legal numbers and illegal
73.      * numbers.  That is the largest legal value, divided by the
74.      * base.  An input number that is greater than this value, if
75.      * followed by a legal input character, is too big.  One that
76.      * is equal to this value may be valid or not; the limit
77.      * between valid and invalid numbers is then based on the last
78.      * digit.  For instance, if the range for longs is
79.      * [-2147483648..2147483647] and the input base is 10,
80.      * cutoff will be set to 214748364 and cutlim to either
81.      * 7 (neg==0) or 8 (neg==1), meaning that if we have accumulated
82.      * a value > 214748364, or equal but the next digit is > 7 (or 8),
83.      * the number is too big, and we will return a range error.
84.      *
85.      * Set any if any `digits' consumed; make it negative to indicate
86.      * overflow.
87.      */ 
88.     cutoff = neg ? -(unsigned long)LONG_MIN : LONG_MAX; 
89.     cutlim = cutoff % (unsigned long)base; 
90.     cutoff /= (unsigned long)base; 
91.     for (acc = 0, any = 0;; c = *s++) { 
92.         if (isdigit(c)) 
93.             c -= '0'; 
94.         else if (isalpha(c)) 
95.             c -= isupper(c) ? 'A' - 10 :'a' - 10; 
96.         else 
97.             break; 
98.         if (c >= base) 
99.             break; 
100.         if (any < 0 || acc > cutoff || acc == cutoff && c > cutlim) 
101.             any = -1; 
102.         else { 
103.             any = 1; 
104.             acc *= base; 
105.             acc += c; 
106.         } 
107.     } 
108.     if (any < 0) { 
109.         acc = neg ? LONG_MIN : LONG_MAX; 
110. //      errno = ERANGE; 
111.     } else if (neg) 
112.         acc = -acc; 
113.     if (endptr != 0) 
114.         *endptr = (char *)(any ? s - 1 : nptr); 
115.     return (acc); 
116. } 
117.  
118. /*
119. * Convert a string to an unsigned long integer.
120. *
121. * Ignores `locale' stuff.  Assumes that the upper and lower case
122. * alphabets and digits are each contiguous.
123. */ 
124. unsigned long 
125. strtoul(nptr, endptr, base) 
126.     const char *nptr; 
127.     char **endptr; 
128.     register int base; 
129. { 
130.     register constchar *s = nptr; 
131.     register unsigned long acc; 
132.     register int c; 
133.     register unsigned long cutoff; 
134.     register int neg = 0, any, cutlim; 
135.  
136.     /*
137.      * See strtol for comments as to the logic used.
138.      */ 
139.     do { 
140.         c = *s++; 
141.     } while (isspace(c)); 
142.     if (c == '-') { 
143.         neg = 1; 
144.         c = *s++; 
145.     } else if (c =='+') 
146.         c = *s++; 
147.     if ((base == 0 || base == 16) && 
148.         c == '0' && (*s == 'x' || *s =='X')) { 
149.         c = s[1]; 
150.         s += 2; 
151.         base = 16; 
152.     } else if ((base == 0 || base == 2) && 
153.         c == '0' && (*s =='b' || *s == 'B')) { 
154.         c = s[1]; 
155.         s += 2; 
156.         base = 2; 
157.     } 
158.     if (base == 0) 
159.         base = c == '0' ? 8 : 10; 
160.     cutoff = (unsigned long)ULONG_MAX / (unsignedlong)base; 
161.     cutlim = (unsigned long)ULONG_MAX % (unsignedlong)base; 
162.     for (acc = 0, any = 0;; c = *s++) { 
163.         if (isdigit(c)) 
164.             c -= '0'; 
165.         else if (isalpha(c)) 
166.             c -= isupper(c) ? 'A' - 10 :'a' - 10; 
167.         else 
168.             break; 
169.         if (c >= base) 
170.             break; 
171.         if (any < 0 || acc > cutoff || acc == cutoff && c > cutlim) 
172.             any = -1; 
173.         else { 
174.             any = 1; 
175.             acc *= base; 
176.             acc += c; 
177.         } 
178.     } 
179.     if (any < 0) { 
180.         acc = ULONG_MAX; 
181. //      errno = ERANGE; 
182.     } else if (neg) 
183.         acc = -acc; 
184.     if (endptr != 0) 
185.         *endptr = (char *)(any ? s - 1 : nptr); 
186.     return (acc); 
187. } 

static inline intisupper(char c){    return (c >= 'A' && c <= 'Z');}static inline intisalpha(char c){    return ((c >= 'A' && c <= 'Z') || (c >= 'a' && c <= 'z'));}static inline intisspace(char c){    return (c == ' ' || c == '\t' || c == '\n' || c == '\12');}static inline intisdigit(char c){    return (c >= '0' && c <= '9');}/* * Convert a string to a long integer. * * Ignores `locale' stuff.  Assumes that the upper and lower case * alphabets and digits are each contiguous. */longstrtol(nptr, endptr, base)const char *nptr;char **endptr;register int base;{register const char *s = nptr;register unsigned long acc;register int c;register unsigned long cutoff;register int neg = 0, any, cutlim;/* * Skip white space and pick up leading +/- sign if any. * If base is 0, allow 0x for hex and 0 for octal, else * assume decimal; if base is already 16, allow 0x. */do {c = *s++;} while (isspace(c));if (c == '-') {neg = 1;c = *s++;} else if (c == '+')c = *s++;if ((base == 0 || base == 16) &&    c == '0' && (*s == 'x' || *s == 'X')) {c = s[1];s += 2;base = 16;} else if ((base == 0 || base == 2) &&    c == '0' && (*s == 'b' || *s == 'B')) {c = s[1];s += 2;base = 2;}if (base == 0)base = c == '0' ? 8 : 10;/* * Compute the cutoff value between legal numbers and illegal * numbers.  That is the largest legal value, divided by the * base.  An input number that is greater than this value, if * followed by a legal input character, is too big.  One that * is equal to this value may be valid or not; the limit * between valid and invalid numbers is then based on the last * digit.  For instance, if the range for longs is * [-2147483648..2147483647] and the input base is 10, * cutoff will be set to 214748364 and cutlim to either * 7 (neg==0) or 8 (neg==1), meaning that if we have accumulated * a value > 214748364, or equal but the next digit is > 7 (or 8), * the number is too big, and we will return a range error. * * Set any if any `digits' consumed; make it negative to indicate * overflow. */cutoff = neg ? -(unsigned long)LONG_MIN : LONG_MAX;cutlim = cutoff % (unsigned long)base;cutoff /= (unsigned long)base;for (acc = 0, any = 0;; c = *s++) {if (isdigit(c))c -= '0';else if (isalpha(c))c -= isupper(c) ? 'A' - 10 : 'a' - 10;elsebreak;if (c >= base)break;if (any < 0 || acc > cutoff || acc == cutoff && c > cutlim)any = -1;else {any = 1;acc *= base;acc += c;}}if (any < 0) {acc = neg ? LONG_MIN : LONG_MAX;//errno = ERANGE;} else if (neg)acc = -acc;if (endptr != 0)*endptr = (char *)(any ? s - 1 : nptr);return (acc);}/* * Convert a string to an unsigned long integer. * * Ignores `locale' stuff.  Assumes that the upper and lower case * alphabets and digits are each contiguous. */unsigned longstrtoul(nptr, endptr, base)const char *nptr;char **endptr;register int base;{register const char *s = nptr;register unsigned long acc;register int c;register unsigned long cutoff;register int neg = 0, any, cutlim;/* * See strtol for comments as to the logic used. */do {c = *s++;} while (isspace(c));if (c == '-') {neg = 1;c = *s++;} else if (c == '+')c = *s++;if ((base == 0 || base == 16) &&    c == '0' && (*s == 'x' || *s == 'X')) {c = s[1];s += 2;base = 16;} else if ((base == 0 || base == 2) &&    c == '0' && (*s == 'b' || *s == 'B')) {c = s[1];s += 2;base = 2;}if (base == 0)base = c == '0' ? 8 : 10;cutoff = (unsigned long)ULONG_MAX / (unsigned long)base;cutlim = (unsigned long)ULONG_MAX % (unsigned long)base;for (acc = 0, any = 0;; c = *s++) {if (isdigit(c))c -= '0';else if (isalpha(c))c -= isupper(c) ? 'A' - 10 : 'a' - 10;elsebreak;if (c >= base)break;if (any < 0 || acc > cutoff || acc == cutoff && c > cutlim)any = -1;else {any = 1;acc *= base;acc += c;}}if (any < 0) {acc = ULONG_MAX;//errno = ERANGE;} else if (neg)acc = -acc;if (endptr != 0)*endptr = (char *)(any ? s - 1 : nptr);return (acc);}

     我们来分析下strtol程序逻辑吧。程序里看到了几个inline函数，这个是实现ctype.h里面字符类型判断

的函数，为后面扫描字符时做准备。使用inline函数的好处是高效，适用于函数代码短小的情况。

register变量存放在寄存器中，CPU对其访问比内存中变量的访问快，适合需要频繁访问的变量。

里面除了定义了用于扫描字串的指针s（指向当前扫描字符的下一个），还定义指向当前扫描字符的c。

（1）程序能处理base为0或者16，而处理前导空格符及正负号后的字串以0x，0X开头的情况（16进制）。

（2）程序能处理base为0或者2，而处理前导空格符及正负号后的字串以0b，0B开头的情况（2进制）。

（3）似乎程序并没有处理base为8，而处理前导空格符及正负号后的字串以'0'开头的情况（8进制）。

事实上，任何进制数的前导0，是”隐式“被处理了的。因为累加和的计算方式是当前的和加上下一个被计入

的字符。如果发现了前导0，扫描后，当前累加和还是0，sum * base + [ 下一个字符数值] =[ 下一个字符数值]

（4）这里有个疑问：这部分程序的目的是想让c被赋值为”处理前导空格，制表符及符号符之后的第一个字符“，

而s指向c后面的字符。如果输入是"0x”或者“0X”怎么办？我们知道，c为'\0'，那s呢？ 此处是否涉及到越界？

（5）看见那个cutoff没？那个可是个unsigned int类型的，所以能够”容纳“INT_MAX或者-INT_MIN。与我

上一篇文章程序中overflow的意图是一样的。cutlim为cutoff除以base所得余数；之后cutoff变为所能达到

最大值的绝对值除以base所得的”商“。看到这个商和余数，联想到了什么？

如果已经扫描到的数已经大于这个“最大商”，那么，接下来就不能出现合法字符（数字，字母），否则溢出。

因为：sum * base +[ 下一个字符数值] > sum * base > 当前符号下所对应的最大数值

如果已经扫描到的数等于这个“最大商”，那么接下来扫描到的合法字符（数字，字母）不能超过此时的“最大余数”。

sum * base +[ 下一个字符数值] < sum * base +  "最大余数“ = 当前符号可表示的最大数的绝对值

如果已经扫描到的数小于这个“最大商”，那么接下来扫描到的合法字符（数字，字母）不能超过此时的“最大余数”。

sum * base +[ 下一个字符数值] < ”最大商“ * base +  "最大余数“ = 当前符号可表示的最大数的绝对值

由此，我们得出了扫描数溢出的条件：已经扫描到的数值(sum) > ”最大商“ 或者 （sum == "最大商” 并且

当前扫描的字符数值 > "最大余数“ 。

（6）看看循环里的条件c = *s++，感觉存在上述疑问（4）中提出的问题：c为结束符'\0'时，s指向哪里？

另外，判断溢出时，条件any < 0 是必要的么？感觉根本不需要。

（7）退出循环后，any < 0 对应着数值溢出的情况。对endptr是否为NULL进行判断，如果未NULL，说明并

未指向有效的内存，当然不能引用。如果any不为0 （上面看到是1），说明已经扫描到了有效字符，s -1 指

向c，对于扫描完所有字符的情况，此时c为'\0' ；对于包含非法字符的情况，则c为找到的第一个非法字符。

（8）错误代码时存放在全局变量errno，与我前一篇定义错误码并作为strtoi函数返回值是不一样的。

    

    大家也可参考另一类似程序：http://blog.csdn.net/ammana_babi/article/details/1473336

    

    在网上闲逛，无意中看到了JULY大神写的一篇关于strtoInt的文章。该文从题目分析开始，先拿出大家

容易写出的”错误程序”。然后深入分析，找出该如何去进行错误检查和处理，如何可以优化程序。除了

点赞，我不知道还能做些啥，分享给大家，希望大家有空一定要去看看。网址：JULY的strtoInt

      

     到此，文章要结束了。大家还记得兄弟篇的第一篇么？ 在此也附上网址。

面试题strtoi实现（一）—— 函数的简单实现