为什么用gcc编译时没有的getaddrinfo被发现和性病= C99 [英] Why can't getaddrinfo be found when compiling with gcc and std=c99

问题描述

我有以下的code，我试图编译。当我与性病= C99试了一下有关式结构addrinfo中的隐式声明和函数的getaddrinfo的隐式声明的警告失败。它与STD = gnu99。

I have the following code which I was trying to compile. When I tried with std=c99 it failed with warnings about "implicit declaration of type struct addrinfo" and "implicit declaration of function getaddrinfo". It works with std=gnu99.

#include <sys/types.h>
#include <sys/socket.h>
#include <netdb.h>

int fails(const char *host, const char *port, struct addrinfo *hints)
{
        int rc;
        struct addrinfo *results;

        // can't find this function??
        rc = getaddrinfo(host, port, hints, &results);

        // free memory in this important application
        freeaddrinfo(results);

        return rc;
}

我用来编译的命令是：

The commands I used to compile is:

gcc -c -o fail.o -Wall -Werror -std=c99 -save-temps fail.c
gcc -c -o fail.o -Wall -Werror -std=gnu99 -save-temps fail.c

看着fail.i（preprocessed头）我看到的编译器是正确的：那些类型没有在头被宣布为拉

Looking at fail.i (preprocessed header) I see that compiler is right: those types haven't been declared in the headers pulled in.

于是我去了头，发现的getaddrinfo被保护的#ifdef __USE_POSIX，这与C99编译时显然没有声明包围。

So I went to the headers and noticed that getaddrinfo is surrounded by a guard #ifdef __USE_POSIX, which is obviously not declared when compiling with c99.

我如何告诉我要使用C99和POSIX GCC？我真的不希望使用gnu99的情况下，我决定以后改用编译器（如锵或ICC）。

How do I tell gcc that I want to use c99 and POSIX? I don't really want to use gnu99 in case I decide to switch compilers later (eg Clang or icc).

推荐答案

很简单，因为的getaddrinfo （POSIX.1g扩展名）不是C99标准的一部分：

Simply because getaddrinfo (POSIX.1g extension) is not part of the standard c99:

<一个href=\"http://www.schweikhardt.net/identifiers.html\">http://www.schweikhardt.net/identifiers.html

连住 -std = gnu99 或 -D_POSIX_C_SOURCE = 200112L

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