Why there is no semicolons after preprocessor directives ?

Question

If I write

#include <stdio.h>;

there no error but a warning comes out during compilation

pari.c:1:18: warning: extra tokens at end of #include directive

What is the reason ?

Answer 1

Because they're unnecessary. Preprocessor directives only exist on one line, unless you explicitly use a line-continuation character (for e.g. a big macro).

Answer 2

Because Preprocessor directives are lines included in the code of our programs that are not program statements but directives for the preprocessor.

These preprocessor directives extend only across a single line of code. As soon as a newline character is found, the preprocessor directive is considered to end. That's why no semicolon (;) is expected at the end of a preprocessor directive.

Answer 3

The reason is that preprocessor directives don't use semicolons. This is because they use a line break to delimit statements. This means that you cannot have multiple directives per line:

#define ABC #define DEF // illegal

But you can have one on multiple lines by ending each line (except the last) with a \ (or /, I forget).

Answer 4

Preprocessor directives are a different language than C, and have a much simpler grammar, because originally they were "parsed", if you can call it that, by a different program called cpp before the C compiler saw the file. People could use that to pre-process even non-C files to include conditional parts of config files and the like.

There is a Linux program called "unifdef" that you can still use to remove some of the conditional parts of a program if you know they'll never be true. For instance, if you have some code to support non-ANSI standard compilers surrounded by #ifdef ANSI/#else/#end or just #ifndef ANSI/#end, and you know you'll never have to support non-ANSI any more, you can eliminate the dead code by running it through unifdef -DANSI.

Answer 5

and if you use #define MACRO(para) fun(para); it could be WRONG to put an semikolon behind it.

if (cond) 
  MACRO (par1);
else
  MACRO (par2);

leads to an syntactical error

Answer 6

During compilation, your code is processed by two separate programs, the pre-processor and the compiler. The pre-processor runs first.

Your code is actually comprised of two languages, one overlaid on top of another. The pre-processor deals with one language, which is all directives starting with "#" (and the implications of these directives). It processes the "#include", "#define" and other directives, and leaves the rest of the code untouched (well, except as side effect of the pre-processor directives, like macro substitutions etc.).

Then the compiler comes along and processes the output generated by the pre-processor. It deals with "C" language, and pretty much ignores the pre-processor directives.

The answer to your question is that "#include" is a part of the language processed by the pre-processor, and in this language ";" are not required, and are, in fact, "extra tokens".