GCC的选择最严格的C代码？

问题描述

GCC选项应该设置为尽可能严格GCC？ （并且我的意思是尽可能严格）我在C89中编写代码，并希望我的代码符合ANSI / ISO标准。

解决方案

我推荐使用：

  -Wall -Wextra -std = c89 -pedantic -Wmissing-prototypes -Wstrict-prototypes \ 
 -Wold-style-definition 
  

您应该使用 -O 以及 -g ，因为一些警告仅在使用优化器时可用（实际上，我通常使用 -O3 来发现问题）。您可能更喜欢 -std = gnu89 ，因为这会禁用库中较少的扩展名。 OTOH，如果你编码严格的ANSI C89，也许你希望他们禁用。 -ansi 选项相当于 -std = c89 ，但不像显式或灵活。

缺少的原型会警告您在使用的函数（或定义的外部函数）时没有范围原型。严格的原型意味着你不能在函数声明或定义（或函数指针）中使用'空括号';你需要（void）或正确的参数列表。旧样式定义点K& R样式函数定义，如：

  int old_style（a，b）int a;双b; {...} 
  

如果幸运的话，您不必担心这一点。我在工作中并不那么幸运，而且我也不能使用严格的原型，对我的懊恼来说，因为周围有太多松散的函数指针，所以我不能使用严格的原型。

另见：

What GCC options should be set to have GCC as strict as possible? (and I do mean as strict as possible) I'm writing in C89 and want my code to be ANSI/ISO compliant.

解决方案

I'd recommend using:

-Wall -Wextra -std=c89 -pedantic -Wmissing-prototypes -Wstrict-prototypes \
    -Wold-style-definition

You should compile with -O as well as -g as some warnings are only available when the optimizer is used (actually, I usually use -O3 for spotting the problems). You might prefer -std=gnu89 as that disables fewer extensions in the libraries. OTOH, if you're coding to strict ANSI C89, maybe you want them disabled. The -ansi option is equivalent to -std=c89 but not quite as explicit or flexible.

The missing prototypes warns you about functions which are used (or external functions defined) without a prototype in scope. The strict prototypes means you can't use 'empty parentheses' for function declarations or definitions (or function pointers); you either need (void) or the correct argument list. The old style definition spots K&R style function definitions, such as:

int old_style(a, b) int a; double b; { ... }

If you're lucky, you won't need to worry about that. I'm not so lucky at work, and I can't use strict prototypes, much to my chagrin, because there are too many sloppy function pointers around.

See also: What is the best command-line tool to clean up code

这篇关于GCC的选择最严格的C代码？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！