C99删除stricmp（）和strnicmp（）？ [英] C99 remove stricmp() and strnicmp()?

问题描述

时的功能 stricmp（）和 strnicmp（） C99中去掉？
我总是得到funtion stricmp预警的隐式声明（）（也是 strnicmp（））当我尝试编译它针对C99。
例如，简单的code下面让我警告。

Is the functions stricmp() and strnicmp() removed in C99? I always get warning implicit declaration of funtion stricmp() (and also strnicmp() ) when I try to compile it against C99. For example, the simple code below get me that warning.

#include<string.h>
#include<stdio.h>

char arr[100]="hello";
char arr2[100]="hEllo";

int main()
{
   int n=-1;
   printf("%d\n",n);
   n=strnicmp(arr,arr2,3);   // the same when use the function stricmp();
   printf("%d\n",n);

   getchar();
   return 0;
}

当我尝试编译这块code对C99（的gcc -Wall -std = C99的main.c -o主），我得到的警告。但是，当我编译它，而不 -std = C99 ，没有警告将被抛出。
然而，即使是隐式声明的警告，我的code还是顺利实施。

When I try to compile this piece of code against C99(gcc -Wall -std=c99 main.c -o main), I get that warning. But when I compile it without the -std=c99, no warning will be thrown. However, even though there is warning of implicit declaration, my code still work out right.

这是为什么？那是一个错误吗？如果没有错误，那么什么是C99的完全变化，这使该警告发生的？

Why is that? Is that a bug? If not a bug, then what is exactly the change of C99 which make that warning happen?

推荐答案

在code。与C99编译，它符合C99标准，它不具有 stricmp（）。当code编译没有C99交换机，它符合一个未知的标准实现 stricmp（）。 （鉴于 GCC 没有 -std = C99 ，可能编译到C89 / 90标准wihich允许隐式声明。）

When code compiles with C99, it conform to the C99 standard, which does not have stricmp(). When code compile without C99 switch, it conforms to an unknown standard that implements stricmp(). (Given gcc without -std=c99, likely compiles to the C89/90 standard wihich allows implicit declarations.)

由于 @Joachim Pileborg 评论，比较不敏感是不是C标准的一部分。

As @Joachim Pileborg commented, insensitive compares are not part of the C standard.

使用C99隐函数需要一个诊断（一个在这种情况下，警告）。如果没有C99，函数的隐式使用不会产生警告。该功能存在的这个的编译器的库 - 它只是一个问题是在使用前声明的函数。

With C99 implicit functions require a diagnostic (a warning in this case). Without C99, the implicit use of the function generates no warning. The functions exists in this compiler's library - it is just a question of are the functions declared before use.

很容易的让自己的：

int wal_stricmp(const char *a, const char *b) {
  int ca, cb;
  do {
     ca = (unsigned char) *a++;
     cb = (unsigned char) *b++;
     ca = tolower(toupper(ca));
     cb = tolower(toupper(cb));
   } while (ca == cb && ca != '\0');
   return ca - cb;
}

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请注意：当编码并试图让 AZ 匹配 AZ ，字符串大小写比较程序往往工作均匀良好。但是，试图向的为了的字符串，事情很快变得不可收拾时。 ABC与_bc来临之前或其他依靠，如果同情做大写或小写之后。 '_'，在ASCII，上和小写字母之间。随着国际和区域问题，情况变得更加复杂。我的code示例使用转换的一个来回，以应对那里的大写字母数字符没有一个1对1的映射与小写字母的问题。 IMO强劲不区分大小写的比较复杂责成​​使用UTF编码和病例定义的。

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Note: When coding and trying to make A-Z match a-z, string insensitive compare routines tend to work uniformly well. But when trying to to order strings, things quickly get out of hand. "abc" vs. "_bc" can come before or after the other depending on if compassion was done as upper or lower case. '_', in ASCII, exists between the upper and lower case letters. With internationalization and locale issues, the situation becomes more complex. My code example uses a round-trip of conversion to cope with issues where the number of uppercase char does not have a 1-to-1 mapping with lowercase ones. IMO the complexities of robust case insensitive compares obliges the use of UTF encoding and its case definition.

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