What is 1LL or 2LL in C and C++?

Question

I was looking at some of the solutions in Google Code Jam and some people used this things that I had never seen before. For example,

2LL*r+1LL

What does 2LL and 1LL mean?

Their includes look like this:

#include <math.h>
#include <algorithm>
#define _USE_MATH_DEFINES

or

#include <cmath>

Kyurem · Accepted Answer · 2013-04-27T04:12:20.003

126

The LL makes the integer literal of type long long.

So 2LL, is a 2 of type long long.

Without the LL, the literal would only be of type int.

This matters when you're doing stuff like this:

1   << 40
1LL << 40

With just the literal 1, (assuming int to be 32-bits, you shift beyond the size of the integer type -> undefined behavior). With 1LL, you set the type to long long before hand and now it will properly return 2^40.

edited Apr 27 '13 at 04:12

answered Apr 27 '13 at 04:07

Kyurem

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`long long` was added to the latest C++ standard, I believe. That said, it doesn't matter if you use uppercase `L` or lowercase `l`. Further, there's `u` or `U` for `unsigned` to combine with them. – Ulrich Eckhardt Apr 27 '13 at 06:30
3

@doomster, You are correct. `long long` is an official type in C++11 with at least 64 bits. – chris Apr 27 '13 at 06:39
7

@doomster: Well, it matters for human readability: `11LL` is easier to read than `11ll`. – Keith Thompson Apr 27 '13 at 08:44
how write 0LL in ruby? – user1735921 Jul 17 '17 at 07:09
The equivalent in Ruby might be to explicitly create a Bignum, rather than using an integer literal (which will give you a Fixnum if small enough or a Bignum otherwise). – Sean Burton Oct 20 '17 at 11:51
What is the difference between ((long long)1<<40) and (1LL<<40) ? – Praveen Ojha Jan 02 '19 at 07:56

What is 1LL or 2LL in C and C++?

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