Question

when assigning a literal to a variable of the byte type, if the literal is too large to be stored as a byte value, it ______. a receives a compile error b causes overflow c cannot happen in java d causes underflow e causes no error

Explanation:

In Java, when assigning a literal to a byte type variable, the byte type has a range (e.g., -128 to 127 for signed bytes). If the literal is outside this range (too large), the compiler will detect this type mismatch and throw a compile error. Overflow (B) happens during arithmetic operations, not direct assignment of an out - of - range literal. Option C is wrong as such a situation can occur. Underflow (D) is for values too small, and E is wrong as a compile error will occur.

Answer:

A. receives a compile error