Strange printf and puts statements explained
c, printf, puts
Solution
`" \n"[i==n]` takes the expression `i==n`, which evaluates to either 0 or 1, and uses it as an index into the array `" \n"`, obtaining either `' '` or `'\n'`.
`"YES\0N0"+condition * 4` takes the array `"YES\0N0"`, which 'decays' to a pointer to its first element when used in most expressions, including this one, and adds `condition * 4` to this pointer. If `condition` is 1, that yields a pointer to the `'N'` at the beginning of `"N0"`.
Problem
I came across these two statements in a SNS photo album, with the title "the most elegant way of output I ever met" or something. Here are the two statements: ``` printf("%d%c", a, " \n"[i==n]); puts("YES\0No"+condition * 4); ``` I've no idea what they are doing and how they work. Will someone explain to me? Thank you.