s of Truncated Values Binary in Java

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Table 53
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Examples of Truncated Values Binary
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0000000000000000000001000000010000101011110100001100001100001111 00101011110100001100001100001111 1100001100001111 00001111 0000000001110000
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Decimal
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4415961481999 735101711 -15601 15 'p' (1) (2) (3) (4) (5)
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The discussion on numeric assignment conversions also applies to numeric parameter values at method invocation (see Section 37, p 82), except for the narrowing conversions, which always require a cast The following examples illustrate boxing and unboxing in assignment context:
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Boolean boolRef = true; Byte bRef = 2; // Byte bRef2 = 257; Integer iRef3 = (short)10; // // // // // boxing constant constant constant widening in range: narrowing to byte, then boxing not in range: cast required in range: casting by narrowing to short, to int, then boxing
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short s = 10; // Integer iRef1 = s; boolean bv1 = boolRef; byte b1 = bRef;
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// narrowing // short not assignable to Integer // unboxing // unboxing
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Integer iRefVal = null; // Always allowed int j = iRefVal; // NullPointerException! if (iRefVal != null) j = iRefVal; // Avoids the exception
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55: THE SIMPLE ASSIGNMENT OPERATOR =
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Review Questions
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51 Given the following declaration:
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char c = 'A';
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What is the simplest way to convert the character value in c into an int Select the one correct answer (a) int i = c; (b) int i = (int) c; (c) int i = CharactergetNumericValue(c); 52 What will be the result of compiling and running the following program
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public class Assignment { public static void main(String[] args) { int a, b, c; b = 10; a = b = c = 20; Systemoutprintln(a); } }
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Select the one correct answer (a) The program will fail to compile since the compiler will report that the variable c in the multiple assignment statement a = b = c = 20; has not been initialized (b) The program will fail to compile, because the multiple assignment statement a = b = c = 20; is illegal (c) The code will compile and print 10, when run (d) The code will compile and print 20, when run 53 What will be the result of compiling and running the following program
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public class MyClass { public static void main(String[] args) { String a, b, c; c = new String("mouse"); a = new String("cat"); b = a; a = new String("dog"); c = b; Systemoutprintln(c); } }
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Select the one correct answer (a) The program will fail to compile (b) The program will print mouse, when run
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CHAPTER 5: OPERATORS AND EXPRESSIONS
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(c) The program will print cat, when run (d) The program will print dog, when run (e) The program will randomly print either cat or dog, when run
56 Arithmetic Operators: *, /, %, +, Arithmetic operators are used to construct mathematical expressions as in algebra Their operands are of numeric type (which includes the char type)
Arithmetic Operator Precedence and Associativity
In Table 54, the precedence of the operators is in decreasing order, starting from the top row, which has the highest precedence Unary subtraction has higher precedence than multiplication The operators in the same row have the same precedence Binary multiplication, division, and remainder operators have the same precedence The unary operators have right associativity, and the binary operators have left associativity
Table 54
Arithmetic Operators Unary Binary
+ Addition * + / -
Subtraction Division Subtraction
% Remainder
Multiplication Addition
Evaluation Order in Arithmetic Expressions
Java guarantees that the operands are fully evaluated from left to right before an arithmetic binary operator is applied If evaluation of an operand results in an error, the subsequent operands will not be evaluated In the expression a + b * c, the operand a will always be fully evaluated before the operand b, which will always be fully evaluated before the operand c However, the multiplication operator * will be applied before the addition operator +, respecting the precedence rules Note that a, b, and c are arbitrary arithmetic expressions that have been determined to be the operands of the operators The evaluation order and precedence rules for arithmetic expressions are illustrated in Example 51 The evaluation of each operand in the expression at (1) results in a call of the operandEval() method declared at (2) The first argument to this method is a number to identify the operand and the second argument is the operand value which is returned by the method The output from the program shows that all three operands were evaluated from left to right and the value of the variable i shows that the precedence rules were applied in the evaluation
56: ARITHMETIC OPERATORS: *, /, %, +, -