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Consider the following method. public double calculate(double x) { return x + 1.5; } The following code segment calls the method calculate in the same class. Double d1 = new Double(7.5); System.out.println(calculate(d1)); What, if anything, is printed when the code segment is executed?A. 8.0
B.8.5
C. 9
D. 9.0
E. Nothing is printed because the code does not compile. The actual parameter d1 passed to calculate is a Double, but the formal parameter x is a double.

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ijeggs

Answer:

E (Nothing is printed because the code will not compile)

Explanation:

There is a difference between Double which is a wrapper class and double which is data type. Changing the line Double d1 = new Double(7.5) to Double d1 = new (7.5) or simple declaring double d1 = 7.5 would have giving a printout of 9.0.

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