复合语句(块)是否被 ANSI C 中的括号表达式包围? [英] Are compound statements (blocks) surrounded by parens expressions in ANSI C?

问题描述

浏览我发现的 Linux 内核源代码 某段代码，其中用括号括起来的语句块被视为表达式a la lisp(或ML)，即一个表达式，其值是最后一个的值声明.

Browsing the Linux kernel sources I found some piece of code where a block of statements surrounded by parenthesis is treated as a expression a la lisp (or ML), that is, an expression which value is the value of the last statement.

例如:

int a = ({
    int i;
    int t = 1;
    for (i = 2; i<5; i++) {
        t*=i;
    }
    t;
});

我一直在看 ANSIC 语法试图弄清楚这段代码如何适合解析树，但我没有成功.

I've been looking at the ANSI C grammar trying to figure out how this piece of code would fit in the parse tree, but I haven't been successful.

那么，有没有人知道这种行为是标准规定的还是只是 GCC 的一个特性?

So, does anybody know if this behaviour is mandated by the standard or is just a peculiarity of GCC?

更新:我已经尝试使用标志 -pedantic 并且编译器现在给我一个警告:

Update: I've tried with the flag -pedantic and the compiler now gives me a warning:

warning: ISO C forbids braced-groups within expressions

推荐答案

它被称为表达式中的支撑组".

It's called "braced-group within expression".

ANSI/ISO C 和 C++ 都不允许，但 gcc 支持.

It's not allowed by ANSI/ISO C nor C++ but gcc supports it.

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