While Boann's answer is correct, there is a complication that applies to the case of constant expressions when they appear in assignment contexts.

Consider the following examples:

  char x;
  x = 'a' + 'b';   // OK
  
  char y = 'a';
  y = y + 'b';     // Compilation error

What is going on? They mean the same thing don't they? And why is it legal to assign an int to a char in the first example?

When a constant expression appears in an assignment context, the Java compiler computes the value of the expression and sees if it is in the range of the type that you are assigning to. If it is, then an implicit narrowing primitive conversion is applied.

• In the first example, 'a' + 'b' is a constant expression, and its value will fit in a char, so the compiler allows the implicit narrowing of the int expression result to a char.

• In the second example, y is a variable so y + 'b' is NOT a constant expression. So even though Blind Freddy can see that the value will fit, the compiler does NOT allow any implicit narrowing, and you get a compilation error saying that an int cannot be assigned to a char.

There are some other caveats on when an implicit narrowing primitive conversion is allowed in this context; see JLS 5.2 and JLS 15.28 for the full details. The latter explains in detail the requirements for a constant expression.

While Boann's answer is correct, there is a complication that applies to the case of constant expressions when they appear in assignment contexts.

Consider the following examples:

  char x = 'a' + 'b';   // OK
  
  char y = 'a';
  char z = y + 'b';     // Compilation error

What is going on? They mean the same thing don't they? And why is it legal to assign an int to a char in the first example?

When a constant expression appears in an assignment context, the Java compiler computes the value of the expression and sees if it is in the range of the type that you are assigning to. If it is, then an implicit narrowing primitive conversion is applied.

• In the first example, 'a' + 'b' is a constant expression, and its value will fit in a char, so the compiler allows the implicit narrowing of the int expression result to a char.

• In the second example, y is a variable so y + 'b' is NOT a constant expression. So even though Blind Freddy can see that the value will fit, the compiler does NOT allow any implicit narrowing, and you get a compilation error saying that an int cannot be assigned to a char.

There are some other caveats on when an implicit narrowing primitive conversion is allowed in this context; see JLS 5.2 and JLS 15.28 for the full details.

The latter explains in detail the requirements for a constant expression. It may not be what you may think. (For example, just declaring y as final doesn't help.)

While Boann's answer is correct, there is a complication that applies to the case of constant expressions when they appear in assignment contexts.

Consider the following examples:

  char x;
  x = 'a' + 'b';   // OK
  
  char y = 'a';
  y = y + 'b';     // Compilation error

What is going on? They mean the same thing don't they?

When a constant expression appears in an assignment context, the Java compiler computes the value of the expression and sees if it is in the range of the type that you are assigning to. If it is, then an implicit narrowing conversion is applied.

• In the first example, 'a' + 'b' is a constant expression, and its value will fit in a char, so the compiler allows the implicit narrowing of the int expression result to a char.

• In the second example, y is a variable so y + 'b' is NOT a constant expression. So even though Blind Freddy can see that the value will fit, the compiler doesn't allow any implicit narrowing, and you get a compilation error saying that an int cannot be assigned to a char.

There are some other caveats on when this implicit narrowing is allowed; see JLS 5.2 and JLS 15.28 for the full details.

While Boann's answer is correct, there is a complication that applies to the case of constant expressions when they appear in assignment contexts.

Consider the following examples:

  char x;
  x = 'a' + 'b';   // OK
  
  char y = 'a';
  y = y + 'b';     // Compilation error

What is going on? They mean the same thing don't they? And why is it legal to assign an int to a char in the first example?

When a constant expression appears in an assignment context, the Java compiler computes the value of the expression and sees if it is in the range of the type that you are assigning to. If it is, then an implicit narrowing primitive conversion is applied.

• In the first example, 'a' + 'b' is a constant expression, and its value will fit in a char, so the compiler allows the implicit narrowing of the int expression result to a char.

• In the second example, y is a variable so y + 'b' is NOT a constant expression. So even though Blind Freddy can see that the value will fit, the compiler does NOT allow any implicit narrowing, and you get a compilation error saying that an int cannot be assigned to a char.

There are some other caveats on when an implicit narrowing primitive conversion is allowed in this context; see JLS 5.2 and JLS 15.28 for the full details. The latter explains in detail the requirements for a constant expression.