Also: "Trick is that i,d are the same for printf but match differently for fscanf/scanf according to C11." It's the same for all versions of C. The difference

Also: "Trick is that i,d are the same for printf but match differently for fscanf/scanf according to C11." It's the same for all versions of C. The difference is that f/scanf interprets 0x__h as hexadecimal and 0__ as octal when %i is used. But we aren't using f/scanf here, so it's entirely a moot point since we're using printf where %d and %i are technically equivalent. It would have been a good/obnoxious trick if they made it 0255 instead, though. The LinkdIn question isn't at fault. I'm not even a C programmer (I dabbled, so I can pretty much fake it. You are more experienced than I by quite a bit), yet the answer seems clear enough to me. The use of char instead of int tipped me off to how I would program this in assembler and what the result would be, and gcc seems to agree when I generate assembler from the C code. #include <cstdio> using namespace std; int main() { int a = 255; char b = 255; int c = 0255; char d = 0255; printf("i:\n--\na = %i\nb = %i\nc = %i\nd = %i\n\n", a, b, c, d); printf("d:\n--\na = %d\nb = %d\nc = %d\nd = %d\n\n", a, b, c, d); return(0); } Output: i: -- a = 255 b = -1 c = 173 d = -83 d: -- a = 255 b = -1 c = 173 d = -83