C语言 为什么stdlib.h 中没有strtoi？
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Why is there no strtoi in stdlib.h?
提问by Eli
I have grown accustomed to strtodand variants.
I am wondering why there is no strtoi shipped with stdlib.h.
Why is it that the integer is left out of this party?
我已经习惯了strtod和变种。我想知道为什么 stdlib.h 没有附带 strtoi。为什么整数被排除在这个党之外？
Specifically I am asking why there is not a version of atoi with the safety features of strtod.
具体来说，我在问为什么没有具有 strtod 安全功能的 atoi 版本。
采纳答案by Wiz
strtol()converts a string to an integer, a long integer but an integer nevertheless. There is atoi()but it should be avoided in most cases due to the fact that it lacks a mechanism for error reporting from invalid input.
strtol()将字符串转换为整数，长整数但仍然是整数。有atoi()但在大多数情况下应该避免，因为它缺乏从无效输入报告错误的机制。
回答by chux - Reinstate Monica
Why is there no strtoi in stdlib.h?
为什么stdlib.h 中没有strtoi？
No critical need.
没有关键需求。
In early C, there was not a standard signed type wider than longand all narrower conversions, like int, could be made from strtol()- as done below.
在早期的 C 中，没有一个标准的有符号类型比 更宽long，所有更窄的转换，比如int，都可以从strtol()- 进行 - 如下所示。
IMO, these and their unsignedcounterparts are now missing C functions and a design shortcoming in the current standard C library (C17/18).
IMO，这些及其unsigned对应项现在缺少 C 函数和当前标准 C 库 (C17/18) 中的设计缺陷。
On many systems, longand inthave the same range and so there is a reduced need for a separate strtoi(). Also the atoi()fills the need for quick and dirty code, but can lack error detection. There also is not a strto_short()nor strto_signchar(), etc.
在许多系统上，long并且int具有相同的范围，因此减少了对单独strtoi(). 也满足atoi()了对快速和脏代码的需求，但可能缺乏错误检测。也没有strto_short()norstrto_signchar()等。
It is fairly easy to create a substitute strtoi(). Simplifications exist.
创建替代品相当容易strtoi()。存在简化。
#include <errno.h>
#include <limits.h>
#include <stdlib.h>
static long strto_subrange(const char *s, char **endptr, int base,
long min, long max) {
long y = strtol(s, endptr, base);
if (y > max) {
errno = ERANGE;
return max;
}
if (y < min) {
errno = ERANGE;
return min;
}
return y;
}
// OP's goal
int strtoi(const char *s, char **endptr, int base) {
#if INT_MAX == LONG_MAX && INT_MIN == LONG_MIN
return (int) strtol(s, endptr, base);
#else
return (int) strto_subrange(s, endptr, base, INT_MIN, INT_MAX);
#endif
}
short strtoshort(const char *s, char **endptr, int base) {
return (short) strto_subrange(s, endptr, base, SHRT_MIN, SHRT_MAX);
}
signed char strtoschar(const char *s, char **endptr, int base) {
return (signed char) strto_subrange(s, endptr, base, SCHAR_MIN, SCHAR_MAX);
}
#include <stdint.h>
int16_t strtoint16(const char *s, char **endptr, int base) {
return (int16_t) strto_subrange(s, endptr, base, INT16_MIN, INT16_MAX);
}
回答by James McNellis
The integer isn't left out of the party: there is strtol, which converts a string to a long, which is an integer type.
整数并没有被排除在外：有strtol将字符串转换long为整数类型的 a 。
回答by Marc K.
This is what I have been using.
这是我一直在使用的。
long long_val;
int int_value;
errno = 0;
long_val = strtol (theString, NULL, 10);
if (errno)
handle_error;
if ((long) someIntMin > long_val || long_val > (long) someIntMax)
handle_invalid;
int_value = (int) long_val;
回答by sarnold
Don't overlook the SEE ALSOsection of your manpages :)
不要忽略SEE ALSO联机帮助页的部分:)
SEE ALSO atof(3), atoi(3), atol(3), strtol(3), strtoul(3)
SEE ALSO atof(3), atoi(3), atol(3), strtol(3), strtoul(3)
You're looking for atoi(3). :)
您正在寻找atoi(3). :)
