这是怎么bracketless / braceless code可能是有效的？ [英] How is this bracketless/braceless code possibly valid?

问题描述

所以我穿过这片code来到

So I came across this piece of code:

%:include <stdio.h>

int main(int argc, char *argv<::>) <%
    printf("C program succesfully running...");
    getchar();

    return 0;
%>

这是一些编译器的bug，或者这是新的东西，因为它没有任何问题运行我还没有发现呢。

Is this some compiler bug or is this something new I have not discovered yet because it is running without any problems.

推荐答案

这code采用C有向图，如下解释：<一href=\"http://en.wikipedia.org/wiki/Digraphs_and_trigraphs#C\">http://en.wikipedia.org/wiki/Digraphs_and_trigraphs#C

This code uses C digraphs, as explained here: http://en.wikipedia.org/wiki/Digraphs_and_trigraphs#C

C标准（ISO / IEC 9899：201X，标点符号§6.4.6：3）具有本就此事说：

The C standard (ISO/IEC 9899:201x, Punctuators §6.4.6:3) has this to say on the matter:

在语言的各个方面，六代币

In all aspects of the language, the six tokens

<: :> <% %> %: %:%:

  
  

的行为，分别相同的六个标记

behave, respectively, the same as the six tokens

[ ] { } # ##

  
  

除了他们的拼写。

except for their spelling.

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