申报范围的开头C89局部变量？

问题描述

我试图做到这一点的ANSI C：

I was trying to do this in ANSI C:

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

这不会在Microsoft 工作（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'

这不工作：

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

现在我在你已经在code座中的变量存在的开头声明变量很多地方阅读。这是一般的ANSI C89？

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?

我发现很多论坛上，人们给这个意见，但我没有看到它写在喜欢的手册。

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要求变量在范围的开始申报。这被放宽C99。

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

这是 GCC明确当您使用 -pedantic 标志，它更紧密地强制执行的标准规则（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).

请注意，虽然，这是有效的C89 code：

Note though, that this is valid C89 code:

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

但使用括号来表示一个范围（因而在该范围变量的寿命）​​的似乎并不特别受欢迎，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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