What do 0LL or 0x0UL mean?
These are constants in C and C++. The suffix LL
means the constant is of type long long
, and UL
means unsigned long
.
In general, each L
or l
represents a long
and each U
or u
represents an unsigned
. So, e.g.
1uLL
means the constant 1 with type unsigned long long
.
This also applies to floating point numbers:
1.0f // of type 'float'
1.0 // of type 'double'
1.0L // of type 'long double'
and strings and characters, but they are prefixes:
'A' // of type 'char'
L'A' // of type 'wchar_t'
u'A' // of type 'char16_t' (C++0x only)
U'A' // of type 'char32_t' (C++0x only)
In C and C++ the integer constants are evaluated using their original type, which can cause bugs due to integer overflow:
long long nanosec_wrong = 1000000000 * 600;
// ^ you'll get '-1295421440' since the constants are of type 'int'
// which is usually only 32-bit long, not big enough to hold the result.
long long nanosec_correct = 1000000000LL * 600;
// ^ you'll correctly get '600000000000' with this
int secs = 600;
long long nanosec_2 = 1000000000LL * secs;
// ^ use the '1000000000LL' to ensure the multiplication is done as 'long long's.
In Google Go, all integers are evaluated as big integers (no truncation happens),
var nanosec_correct int64 = 1000000000 * 600
and there is no "usual arithmetic promotion"
var b int32 = 600
var a int64 = 1000000000 * b
// ^ cannot use 1000000000 * b (type int32) as type int64 in assignment
so the suffixes are not necessary.
0LL
is a long long zero.
0x0UL
is an unsigned long zero, expressed using hexadecimal notation. 0x0UL
== 0UL
.
There are several different basic numeric types, and the letters differentiate them:
0 // normal number is interpreted as int
0L // ending with 'L' makes it a long
0LL // ending with 'LL' makes it long long
0UL // unsigned long
0.0 // decimal point makes it a double
0.0f // 'f' makes it a float