Question

I’m confused why Java integer literals default to int instead of long. This seems to cause unnecessary confusion.

First, it requires the programmer to adopt a special syntax (append “L” to literals) when assigning a value to a long that exceeds the maximum int size (2147483647).

long x = 2147483647; // Compiles
long y = 2147483648; // Does not compile
long z = 2147483648L; // Compiles

Second, when using the Long wrapper class, the programmer must always use the long literal notation as explained in this SO question.

Long x = 250; // Does not compile
Long y = 250L; // Compiles

Third, considering that implicit conversion from int literals to the “narrower” data types (short and byte) works just fine in all situations (that I know of), it seems that simply making all integer literals type long would have been the obvious solution… right? Wouldn’t this completely remove the need for this odd system of appending “L” to integer literals in special cases?

Answer 1

This behavior is by design¹ and is codified in the JLS: Java Language Specification.

First, note that this is not related to widening which is why the (valid) integer-literal is promoted to a long value. Instead, this is related to the very specification of the int literal:

It is a compile-time error if a hexadecimal, octal, or binary int literal does not fit in 32 bits.

The smallest and largest signed 32-bit integer values are -2147483648 and 2147483647, respectively.

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¹I care not speculate on why it works this way, and languages like C# have different rules.