I teach Java, I read Shildt. Already many times read that Java - strongly typed language. Why does the following code throw an error

int i = 10; i = i / 2.5;

Does this code work fine and assign a value of 4 to i ?

 int i = 10; i /= 2.5;

    2 answers 2

    This is how an assignment statement works, if a variable is of type int , and i / 2.5 type double , then you cannot assign a value of type double to type int without an explicit cast.

    That is, type checking is performed before assigning a variable to a value. In the second case, the statement is an expression in which the value is converted to the type of the operand , that is, so that the operation can be carried out and back to the type of the variable where the value should be stored. I.e

     int i = 10; d = i /= 2.5;

    will also work, since int values ​​are converted to double before assignment without loss of accuracy, the reverse is not true and causes an error.

    • Does this automatic cast of double to int of the concept of strong typing contradict? - daniilkk
    • The type of the variable does not change during compilation, the type of the variable may change at runtime, however you cannot change the type without casting if this type does not support default type conversion. - Roman C

    type operation

     i /= 2.5;

    this is an automatic type casting operation that will be written as

     i = (int) i/2.5;

    same with

     double d = 1d; int i = 1; i = i * d; // ошибка i *= d; // нет ошибки

    with increment and decrement operations the same situation.