If operators with the same precedence are next to each other, there associativity determines the order of evaluation. All binary operators except assignment operators are left-associative.
Assignment operators are right-associative. Therefore, the expression
a = b += c = 5 equivalent a = (b += (c = 5))

If no operands hava side effects that change the value of a variable, the order of operand evaluation is irrelevant. Interesting cases arise when operands do hava a side effect. For example,

public class TestDemo {
public static void main(String arg[]) {
int a = 0;
int x = ++a + a;
a = 0;
x = a + (++a);

The output will be:

The order for evaluation operands takes precedence over the operator precedence rule. In the former case, (++a) has higher precedence than addition (+), but since a is a left-hand operand of the addition (+), it is evaluated before any part of its right-hand operand (e.g., ++a in this case).

Converting Strings to Numbers:
int intValue = Integer.parseInt(intString);
double doubleValue = Double.parseDouble(doubleString);


public static int parseInt(String s)
throws NumberFormatException
Parses the string argument as a signed decimal integer. The characters in the string must all be decimal digits, except that the first character may be an ASCII minus sign '-' ('\u002D') to indicate a negative value. The resulting integer value is returned, exactly as if the argument and the radix 10 were given as arguments to the parseInt(java.lang.String, int) method.
s - a String containing the int representation to be parsed
the integer value represented by the argument in decimal.

NumberFormatException - if the string does not contain a parsable integer.

Format to keep two digits after the decimal point:
double doubleValue = (int)(doubleValue2 * 100) / 100.0

Computing ab:
public static double pow(double a, double b)

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1.21 Java notes

Posted on 2007-04-22 20:22 ZelluX 阅读(98) 评论(0)  编辑  收藏 所属分类: OOP
2007-01-21 21:57:20