What does accesing a numeric literal with brackets means?
Hello, I spotted this at an International Obfuscated C Code Contest entry, and I don't understand what it means.
The question is: What the heck does something like this means and how the heck does it compiles?:
Code:#include<iostream> using namespace std; int main() { char *a; cout<<35[a]<<endl; //This is the line I don't get at all return 0; }
p[i] = *(p+i);
p +i = i +p;
so p[i] = i[p];
c-faq
Oh, I see. I didn't though that would be on the FAQ, so I didn't searched it hard enough. My bad
Thanks!
It's still simple actually
Code:0[ strchr(p,'\n') ] = '\0'; // combine with function !
Until strchr returns NULL
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
lol, that's just an example!
lol @Salem... why are you always such a purist
1. Get rid of gets(). Never ever ever use it again. Replace it with fgets() and use that instead.
2. Get rid of void main and replace it with int main(void) and return 0 at the end of the function.
3. Get rid of conio.h and other antiquated DOS crap headers.
4. Don't cast the return value of malloc, even if you always always always make sure that stdlib.h is included.
