NAME
strtoul, strtoull, strtouq - convert a string to an unsigned
long integer
SYNOPSIS
#include <stdlib.h>
unsigned long int
strtoul(const char *nptr, char **endptr, int base);
unsigned long long int
strtoull(const char *nptr, char **endptr, int base);
DESCRIPTION
The strtoul() function converts the initial part of the string
in nptr to an unsigned long integer value according to the given
base, which must be between 2 and 36 inclusive, or be the spe-
cial value 0.
The string may begin with an arbitrary amount of white space (as
determined by isspace(3)) followed by a single optional '+' or
'-' sign. If base is zero or 16, the string may then include a
'0x' prefix, and the number will be read in base 16; otherwise,
a zero base is taken as 10 (decimal) unless the next character
is '0', in which case it is taken as 8 (octal).
The remainder of the string is converted to an unsigned long int
value in the obvious manner, stopping at the first character
which is not a valid digit in the given base. (In bases above
10, the letter 'A' in either upper or lower case represents 10,
'B' represents 11, and so forth, with 'Z' representing 35.)
If endptr is not NULL, strtoul() stores the address of the first
invalid character in *endptr. If there were no digits at all,
strtoul() stores the original value of nptr in *endptr (and
returns 0). In particular, if *nptr is not '\0' but **endptr is
'\0' on return, the entire string is valid.
The strtoull() function works just like the strtoul() function
but returns an unsigned long long integer value.
RETURN VALUE
The strtoul() function returns either the result of the conver-
sion or, if there was a leading minus sign, the negation of the
result of the conversion represented as an unsigned value,
unless the original (non-negated) value would overflow; in the
latter case, strtoul() returns ULONG_MAX and sets the global
10, the letter 'A' in either upper or lower case represents 10,
'B' represents 11, and so forth, with 'Z' representing 35.)
If endptr is not NULL, strtoul() stores the address of the first
invalid character in *endptr. If there were no digits at all,
strtoul() stores the original value of nptr in *endptr (and
returns 0). In particular, if *nptr is not '\0' but **endptr is
'\0' on return, the entire string is valid.
The strtoull() function works just like the strtoul() function
but returns an unsigned long long integer value.
RETURN VALUE
The strtoul() function returns either the result of the conver-
sion or, if there was a leading minus sign, the negation of the
result of the conversion represented as an unsigned value,
unless the original (non-negated) value would overflow; in the
latter case, strtoul() returns ULONG_MAX and sets the global
variable errno to ERANGE. Precisely the same holds for str-
toull() (with ULLONG_MAX instead of ULONG_MAX).
ERRORS
EINVAL (not in C99) The given base contains an unsupported
value.
ERANGE The resulting value was out of range.
The implementation may also set errno to EINVAL in case no con-
version was performed (no digits seen, and 0 returned).
NOTES
Since strtoul() can legitimately return 0 or LONG_MAX (LLONG_MAX
for strtoull()) on both success and failure, the calling program
should set errno to 0 before the call, and then determine if an
error occurred by checking whether errno has a non-zero value
after the call.
In locales other than the "C" locale, other strings may be
accepted. (For example, the thousands separator of the current
locale may be supported.)
2007-10-26 19:18
strtol()详解（转载）
今天，在review 一些代码的时候，看到了strtol()这个函数，由于以前使用它的时候，还没有深刻的了解，这次，我决定探个究竟。
网上关于这个函数的资料大都来源于同份资料，linux库函数，讲的不够细致。于是，我花了几个小时，认真地尝试其功能，并整理了这篇文章，希望能对C语言的爱好者一些帮助。
希望大家能够将本文中发现的错误及时反馈给我，以便修正。我的信箱是。
+----------------+
| strtol |
+----------------+
i.e. string to long
long int strtol(const char *nptr, char **endptr, int base)
strtol()会将nptr指向的字符串，根据参数base，按权转化为long int, 然后返回这个值。
参数base的范围为2~36,和0；它决定了字符串以被转换为整数的权值。
可以被转换的合法字符依据base而定，举例来说，当base为2时，合法字符为‘0’，‘1’；base为8时，合法字符为‘0’，‘1’，……‘7’；base为10时，合法字符为‘0’，‘1’，……‘9’；base 为16时，合法字符为‘0’，‘1’，……‘9’，‘a’，……‘f’；base为24时，合法字符为‘0’，……‘9’，‘a’，……‘n’，base为36时，合法字符为‘0’，……‘9’，‘a’，……‘z’；等等。其中，不区分大小写，比如，‘A’和‘a’会都会被转化为10。
当字符合法时，‘0’，……‘9’依次被转换为十进制的0～9，‘a’，……‘z’一次北转换为十进制的10～35。
strtol()函数检测到第一个非法字符时，立即停止检测，其后的所有字符都会被当作非法字符处理。合法字符串会被转换为long int, 作为函数的返回值。非法字符串，即从第一个非法字符的地址，被赋给*endptr。**endptr是个双重指针，即指针的指针。strtol()函数就是通过它改变*endptr的值，即把第一个非法字符的地址传给endptr。
多数情况下，endptr设置为NULL, 即不返回非法字符串。
下面看几个例子:
------------------------------------------------------
char buffer[20]="10379cend$3";
char *stop;
printf("%d\n",strtol(buffer, &stop, 2));
printf("%s\n", stop);
输出结果：
2
379cend$3
-------------------------------------------------------
char buffer[20]="10379cend$3";
char *stop;
printf("%d\n",strtol(buffer, &stop, 8));
printf("%s\n", stop);
输出结果：
543
9cend$3
--------------------------------------------------------
char buffer[20]="10379cend$3";
char *stop;
printf("%d\n",strtol(buffer, &stop, 10));
printf("%s\n", stop);
输出结果：
10379
cend$3
-------------------------------------------------------
char buffer[20]="10379cend$3";
char *stop;
printf("%d\n",strtol(buffer, &stop, 16));
printf("%s\n", stop);
输出结果：
17005006
nd$3
另外，如果base为0，且字符串不是以0x(或者0X)开头，则按十进制进行转化。如果base为0或者16，并且字符串以0x（或者0X）开头，那么，x（或者X）被忽略，字符串按16进制转化。如果base不等于0和16，并且字符串以0x(或者0X)开头，那么x被视为非法字符。
例如:
-------------------------------------------------------
char buffer[20]="0x31da6c";
char *stop;
printf("%d\n",strtol(buffer, &stop, 0));
printf("%s\n", stop);
输出结果(stop为空)：
3267180
-------------------------------------------------------
char buffer[20]="0x31da6c";
char *stop;
printf("%d\n",strtol(buffer, &stop, 13));
printf("%s\n", stop);
输出结果：
0
0x31da6c
-------------------------------------------------------
最后，需要说明的是，对于nptr指向的字符串，其开头和结尾处的空格被忽视，字符串中间的空格被视为非法字符。
例如：
-------------------------------------------------------
char buffer_1[20]="10379c";
char buffer_2[20]=" 10379c ";
char buffer_3[20]=" 10 379c ";
printf("%d\n",strtol(buffer_1,NULL,0));
printf("%d\n",strtol(buffer_2,NULL,0));
printf("%d\n",strtol(buffer_3,NULL,0));
输出结果为：
10379
10379
10
--------------------------------------------------------
