long a = 1 和 long a = 1L 的区别

Question

My question is long a = 1; an integer in reality? And is it the same as long a = 1L; in data type?

Answer 1

In Java the default type for integer number literals is int , that means all numeric integer constants are treated as int . If you assign an int literal to a long typed variable, the type of the variable will not change.

If you assign long a = 1L , you're telling the compiler that the literal 1 should be treated as a long .

long a = 3000000000;   // <-- won't work; 3,000,000,000 > Integer.MAX_VALUE

long a = 3000000000L;  // <-- will work; fits into a long

Answer 2

a is still a long , 1 is implicitly converted (only works in this direction, not when shrinking where an implicit cast is required).

You can confirm that it can hold values larger than an integer could:

    public static void main(String[] args) {
        long a = 1;
        a += Integer.MAX_VALUE;
        System.out.println(a); // outputs 2147483648 rather than -2147483648
    }

Answer 3

You can see from the following experiments that the literals (not prefixed with L ) are int s, but that there are automatic casts such that a long variable finaly always holds a long value:

long a = 2147483647 + 1;

long b = 2147483647;
b++;

long c = 2147483647L + 1;

• For a , the literals are int s. The calclulation is done with int s, which overflows. Therefore long a gets assigned -2147483648L .
• For b , the int literal is casted to long during assignment, which can be incremented to the final value b = 2147483648L without overflow.
• For c , the calculation is already done with long s (the int literal 1 is automatically casted to 1L ), giving c = 2147483648L in first place.

You can find the specification for numeric promotion in JLS 5.6 "Numeric Contexts"