When is numeric suffix L (long) needed?

Question

I know that something like 1.5 is a double, so you need to use the m suffix to assign it to a decimal.

decimal d = 1.5; // won't compile
decimal d = 1.5m; // will compile

Now, a literal like 0 is an int. But why can I just assign it to a long? Is there any difference between assigning 0 versus 0L to a variable of type long?

long l = 0;
long l = 0L;

Answer 1

But why can I just assign it to a long?

11.2.3 of the spec states:

Implicit numeric conversions
The implicit numeric conversions are:
•   From sbyte to short, int, long, float, double, or decimal.
•   From byte to short, ushort, int, uint, long, ulong, float, double, or decimal.
•   From short to int, long, float, double, or decimal.
•   From ushort to int, uint, long, ulong, float, double, or decimal.
•   From int to long, float, double, or decimal.
•   From uint to long, ulong, float, double, or decimal.
•   From long to float, double, or decimal.
•   From ulong to float, double, or decimal.
•   From char to ushort, int, uint, long, ulong, float, double, or decimal.
•   From float to double.
Conversions from int, uint, long, or ulong to float and from long or ulong to double may cause a loss of precision, but will never cause a loss of magnitude. The other implicit numeric conversions never lose any information.

From int to long, float, double, or decimal. is the relevant one here. 0 is int by default, so it can be implicitly converted to long.

Is there any difference between assigning 0 versus 0L to a variable of type long?

They will behave the exact same. The IL is the exact same.