把一个字符串转成double类型的数[# 61]

问题：

给一个字符串，比如“-12.05”，把它转成相应的double类型的数。

分析：

在进行转换的时候，要注意以下问题：

1. 该字符串是否为空

2. 是否该字符串含有符号；

3. 该字符串内是否有非法字符；

4. 小数点的位置；

5. 该数是否越界；

代码如下：

public static double atod(String str) throws Exception {
	
	boolean negative = false;
	//get the value before the "."
	double valueBeforeDot = 0.0d;
	//get the value after the ".";
	double valueAfterDot = 0.0d;
	boolean pointAppear = false;
	int count = 0;
	
	//null or empty string
	if (str == null || str.equals("")) {
		throw new Exception("null string or the string has no character!");
	} 
	
	for (int i = 0; i < str.length(); i++) {
		//check whether the first character is "+" or "-"
		if (i == 0 && (str.charAt(0) == '-' || str.charAt(0) == '+')) {
			if (str.charAt(0) == '-') {
				negative = true;				
			}
		} else {
			//check whether the character is "." and appears for 
			//the first time and appears at the correct position.
			if (pointAppear == false && i != 0 && str.charAt(i) == '.' 
					&& (str.charAt(0) != '-' || str.charAt(0) != '+')) {
				pointAppear = true;
			} else {
				if (str.charAt(i) >= '0' && '9' >= str.charAt(i)) {
					if (pointAppear == false) {
						valueBeforeDot = valueBeforeDot * 10 + (str.charAt(i) - '0');
						if (valueBeforeDot > Double.MAX_VALUE) {
							throw new Exception("out of Double range");
						}
					} else {
						valueAfterDot = valueAfterDot * 10 + (str.charAt(i) - '0');
						count++;
					}
				} else {
					throw new NumberFormatException("not a double");
				}
			} 
		}				
	}
	valueBeforeDot = valueBeforeDot + valueAfterDot /Math.pow(10, count);
	return negative == true ? valueBeforeDot * -1 : valueBeforeDot; 			
}

扩展：

把一个字符串转成一个整数。 解答可以参考http://blog.csdn.net/beiyeqingteng/article/details/7000034

转载请注明出处：http://blog.csdn.net/beiyeqingteng