MAP_ANONYMOUS与C99标准 [英] MAP_ANONYMOUS with C99 standard

问题描述

我有一个使用mmap系统调用的应用程序，我在得到它的编译看，为什么我是越来越MAP_ANON和MAP_ANONYMOUS是未申报时间的问题，我不得不code的一小部分，我使用我看到我可以编译它只是罚款，所以我尝试只是一个基本的编译和工作，我看到，当您添加-std = C99失败。是否有一个具体的原因，MAP_ANON和MAP_ANONYMOUS是无效的C99标准？我知道他们不是POSIX定义，但通过BSD源代码的定义，所以我只是想知道这是为什么。

I have an application that uses the mmap system call, I was having an issue getting it to compile for hours looking as to why I was getting MAP_ANON and MAP_ANONYMOUS were undeclared, I had a smaller section of code that I used and I saw I could compile it just fine so I tried just a basic compile and that worked, I saw that it fails when you add -std=c99. Is there a specific reason that MAP_ANON and MAP_ANONYMOUS are not valid in the C99 standard? I know that they aren't defined by POSIX but are defined by BSD SOURCE so I just want to know why that is.

推荐答案

您可能希望 -std = gnu99 而不是 -std = C99 。 C99模式明确禁用（大多数）GNU扩展。

You probably want -std=gnu99 instead of -std=c99. C99 mode explicitly disables (most) GNU extensions.

我写了一个简单的测试：

I wrote a simple test:

#include <sys/mman.h>

int a = MAP_ANONYMOUS;

在C99模式下，它没有找到该值：

In C99 mode, it doesn't find the value:

$ gcc -std=c99 -c d.c
d.c:3:9: error: ‘MAP_ANONYMOUS’ undeclared here (not in a function)

而在Gnu99模式，它的作用：

Whereas in Gnu99 mode, it does:

$ gcc -std=gnu99 -c d.c

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