在旧 C 中使用前声明变量

发布于 2024-12-06 05:51:22 字数 136 浏览 0 评论 0原文

最近，我必须修改使用非常旧的 GCC 版本（版本 2.3 左右）编译的遗留代码。在函数中，变量必须在使用之前声明。我相信这是C89标准所做的。此限制后来被删除。

我的问题是：当时他们为什么要执行这个裁决？是否存在任何可能危及软件完整性的问题？

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萌面超妹 2024-12-13 05:51:22

变量在使用之前仍然需要声明——而且它们从来不需要在函数的顶部声明。

C89 要求块由开始的 {、后跟零个或多个声明、后跟零个或多个语句、后跟结束的 } 组成。

例如，这是合法的 C89（并且，没有 void，甚至是 K&RC，可以追溯到 1978 年或更早）：

int foo(void) {
    int outer = 10;
    {
         int inner = 20;
         printf("outer = %d, inner = %d\n", outer, inner);
    }
    printf("outer = %d, inner is not visible\n", outer);
    return 0;
}

一点，允许声明和语句在块内混合：

int foo(void) {
    int x = 10;
    printf("x = %d\n", x);
    int y = 20;
    printf("y = %d\n", y);
    return 0;
}

C99 放宽了这对于最初的限制，我认为它可以追溯到 C 的祖先语言：B、BCPL，甚至 Algol。它可能确实使编译器的工作变得更容易一些。（我以为这会让解析变得更容易，但我不认为它会这样做；它仍然必须能够在事先不知道上下文的情况下区分某些东西是声明还是语句。）

Variables still have to be declared before being used -- and they've never had to be declared just at the top of a function.

The C89 requirement is that a block consists of an opening {, followed by zero or more declarations, followed by zero or more statements, followed by the closing }.

For example, this is legal C89 (and, without the void, even K&R C, going back to 1978 or earlier):

int foo(void) {
    int outer = 10;
    {
         int inner = 20;
         printf("outer = %d, inner = %d\n", outer, inner);
    }
    printf("outer = %d, inner is not visible\n", outer);
    return 0;
}

C99 loosened this, allowing declarations and statements to be mixed within a block:

int foo(void) {
    int x = 10;
    printf("x = %d\n", x);
    int y = 20;
    printf("y = %d\n", y);
    return 0;
}

As for the reason for the original restriction, I think it goes back to C's ancestor languages: B, BCPL, and even Algol. It probably did make the compiler's job a bit easier. (I was thinking that it would make parsing easier, but I don't think it does; it still has to be able to distinguish whether something is a declaration or a statement without knowing in advance from the context.)

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