把一个字符串转成double类型的数[# 61]

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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 stringif (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