C编译器如何解析以下C语句？ [英] How does the C compiler parse the following C statement?

问题描述

请考虑以下行：

int i;
printf("%d",i);

词法分析器是否会进入字符串来解析％和 d 作为单独的令牌，还是将％d解析为一个令牌？

Will the lexical analyzer go into the string to parse % and d as separate tokens, or will it parse "%d" as one token?

推荐答案

有两个解析器在这里工作：第一，C编译器，将解析C文件，基本上忽略字符串的内容（虽然现代C编译器将解析字符串，以帮助捕获错误的格式字符串 - ％转换说明符与传递给要转换的 printf（） 。

There are two parsers at work here: first, the C compiler, that will parse the C file and basically ignore the content of the string (though modern C compilers will parse the string as well to help catch bad format strings — mismatches between the % conversion specifier and the corresponding argument passed to printf() to be converted).

下一个解析器是内置到C运行时库中的字符串格式解析器。当调用 printf 时，将在运行时调用此函数来解析格式字符串。这个解析器当然比较简单。

The next parser is the string format parser built into the C runtime library. This will be called at runtime to parse the format string when you call printf. This parser is of course very simple in comparison.

我没有检查，但我猜想C编译器，帮助检查错误的格式字符串将实现一个 printf 类解析器作为后处理步骤（即使用自己的词法分析器）。

I have not checked, but I would guess that the C compilers that help checking for bad format strings will implement a printf-like parser as a post-processing step (i.e. using its own lexer).

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