"Overflow" compiler error with -9223372036854775808L

Question

The range of the Long data type is -9223372036854775808 to 9223372036854775807, but the following statement generates compiler error "BC30036: Overflow":

Dim a As Long = -9223372036854775808L

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Why is this an error? How can I specify the constant -9223372036854775808 in code?

Answer 1

Why is this an error?

The compiler parses the expression -9223372036854775808L as a unary minus operator applied to the decimal integer literal 9223372036854775808L. According to the VB.NET specification:

A decimal integer literal is a string of decimal digits (0-9).

And:

If an integer literal's type is of insufficient size to hold the integer literal, a compile-time error results.

9223372036854775808L is too large for a Long, so you get an overflow error. (The minus sign isn't part of the integer literal.)

How can I specify the constant -9223372036854775808 in code?

To specify -9223372036854775808 literally, use a hexadecimal literal:

Dim a As Long = &H8000000000000000

The VB.NET specification alludes to this as well:

Decimal literals directly represent the decimal value of the integral literal, whereas octal and hexadecimal literals represent the binary value of the integer literal (thus, &H8000S is -32768, not an overflow error).

Of course, for clarity you should probably just use Long.MinValue instead of a literal:

Dim a As Long = Long.MinValue

What about C#?

As René Vogt pointed out, the equivalent statement compiles fine in C#:

long a = -9223372036854775808L;

That's because (unlike VB.NET) C# supports this as a special case:

When a decimal_integer_literal with the value 9223372036854775808 (2^63) and no integer_type_suffix or the integer_type_suffix L or l appears as the token immediately following a unary minus operator token, the result is a constant of type long with the value -9223372036854775808 (-2^63). In all other situations, such a decimal_integer_literal is of type ulong.