在作用域的开头声明C89局部变量? [英] Declare C89 local variables in the beginning of the scope?

问题描述

我试图在 ANSI C 中做到这一点:

I was trying to do this in ANSI C:

include <stdio.h>
int main()
{
    printf("%d", 22);
    int j = 0;
    return 0;
}

这在 Microsoft Visual C++ 2010 中不起作用(在 ANSI C项目).你得到一个错误:

This does not work in Microsoft Visual C++ 2010 (in an ANSI C project). You get an error:

error C2143: syntax error : missing ';' before 'type'

这确实有效:

include <stdio.h>
int main()
{
    int j = 0;
    printf("%d", 22);
    return 0;
}

现在我在很多地方读到你必须在变量存在的代码块的开头声明变量.对于ANSI C89来说这通常是真的吗?

Now I read at many places that you have to declare variables in the beginning of the code block the variables exist in. Is this generally true for ANSI C89?

我在很多论坛上找到了人们提供此建议的地方，但我没有看到它以任何官方"来源编写，例如 GNU C 手册.

I found a lot of forums where people give this advice, but I did not see it written in any 'official' source like the GNU C manual.

推荐答案

ANSI C89 要求在作用域的开始处声明变量.这在 C99 中变得轻松.

ANSI C89 requires variables to be declared at the beginning of a scope. This gets relaxed in C99.

当您使用 -pedantic 标志时，gcc 很清楚这一点，它更严格地执行标准规则(因为它默认为 C89 模式).

This is clear with gcc when you use the -pedantic flag, which enforces the standard rules more closely (since it defaults to C89 mode).

但请注意，这是有效的 C89 代码:

Note though, that this is valid C89 code:

include <stdio.h>
int main()
{
    int i = 22;
    printf("%d
", i);
    {
        int j = 42;
        printf("%d
", j);
    }
    return 0;
}

但是使用大括号来表示范围(以及该范围内变量的生命周期)似乎并不特别流行，因此 C99 ......等

But use of braces to denote a scope (and thus the lifetime of the variables in that scope) doesn't seem to be particularly popular, thus C99 ... etc.

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