为什么没有分号会给出错误，而太多的分号却不会? [英] Why no semicolon gives errors but too many of them don't?

问题描述

考虑这个 C 代码:

#include ;int main(void) {puts("你好，世界！");;;;返回0；;;;};

在这里，我几乎在所有可能的地方都放了分号.只是为了乐趣.但令人惊讶的是它奏效了！在包含之后我收到了关于分号的警告，但其他绝对错误的分号起作用了.如果我忘记在 puts 后面放一个分号，我会得到以下错误

<块引用>

错误:应为';'在'返回'之前

<小时>

为什么很多错误和无用的分号不会导致错误?在我看来，它们应该被视为语法错误.

解决方案

单个分号构成一个空语句.它不仅合法，而且在某些情况下也很有用，例如，不需要实体的 while/for 循环.一个例子:

while (*s++ = *t++);

<块引用>

C11 6.8.3 表达式和空语句

空语句(仅由分号组成)不执行任何操作.

<小时>

唯一的语法错误是这一行:

#include ;

Consider this C code:

#include <stdio.h>;

int main(void) {
    puts("Hello, world!");; ;
    ;
    return 0; ;
    ; ;
};

Here I've put semicolons almost everywhere possible. Just for fun. But surprisingly it worked! I got a warning about the semicolon after include but other absolutely wrong semicolons worked. If I forget to put a semicolon after puts, I'll get the following error

error: expected ';' before 'return'

---

Why don't lots of wrong and useless semicolons cause errors? To my mind they should be treated as syntax errors.

解决方案

A single semicolon constructs a null statement. It's not only legal, it's also useful in some cases, for instance, a while/ for loop that doesn't need a real body. An example:

while (*s++ = *t++)
    ;

C11 6.8.3 Expression and null statements

A null statement (consisting of just a semicolon) performs no operations.

---

The only syntax error is this line:

#include <stdio.h>;

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