Tool for braceless, whitespace sensitive C syntax

Question

I\'m writing some C at the moment and because I like whitespace sensitive syntax, I\'d like to write it like this:

#include 

int main(void)
          


        

Answer 1

If you really want to do this, it is not going to be possible without implementing a language parser, and even then, I am not sure how the coding convention will be for some of the cases in your "new language that looks like C but has no braces". For example, take the following C code:

struct a {
    int i;
};

int main(void) {
    ...
}

You can write it as

struct a
    int i

int main(void)
    ...

But it has to be converted to the original code, not:

struct a {
    int i;
} /* Note the missing semicolon! */

int main(void) {
    ...
}

Also, given the snippets below:

/* declare b of type struct a */
struct a {
    int i;
} b;

/* a struct typedef */
typedef struct a {
    int i;
} b;

How are you going to specify these in your language?

You seem to not want to use semicolons in your language either. This restricts your code quite a bit, and makes the conversion tool complicated as well, because you can't have continuation lines without extra effort:

i = j +
k;

is legal C, but

i = j + ;
k;

is not.

So first, you need to define the grammar of your "braceless C" more precisely. As others have said, this sort of thing is fraught with peril.