Comparing long long with 0

Question

long long llIdx = foo();
if (llIdx > 0LL) // Can I use 0 here?
  ...

Is there any problem if I use 0 instead of 0LL in above code?

When should I prefer 0LL over 0?

Solution

Yes, you can use a plain 0 here. The compiler would look at the type of each argument to > and promote the smaller one so that they are the same size.

Thus llIdx > 0 and llIdx > 0LL are equivalent.

Comments

[kay]

long long is a signed type. It makes no difference if you write x > 0ll or x > 0, because 0 is automatically promoted to a long long.

[kay]

"The compiler would look at the type of each argument to > and promote the smaller one so that they are the same size." That's why I didn't put my comment as an answer. There is a bunch of special cases if the sign of both integers vary.

[kdopen]

Didn't see your comment while I was typing :) I was just trying to answer the specific question with a little more than "yes". And I presume you meant "signedness", not "sign"

[Deqing]

So when we should use 0LL instead of 0?

[mafso]

@Deqing: I can't come up with a good example where usual arithmetic conversion does something unexpected if one operand has value 0. Variadic arguments are one (e.g. printf("%lld", 0LL);), and unsigned n; int i; ... long long l = 0LL + n + i; (if long long covers the range of unsigned int, what it probably always does, this prevents promotion of i to unsigned).