【发布时间】:2015-04-18 18:47:51
【问题描述】:
问题:当我尝试将 int 转换为 double 时,它显示 int 无法解析为 variable 的错误。该程序将输入二次方程作为输入,并以这种格式提取 aX2-bX-c=0 的系数并求解二次方程。但是从 int 到 double 的转换是一些错误。
程序:
public static String quad (final String equation)
{
final String regex = "([+-]?\\d+)X2([+-]\\d+)X([+-]\\d+)=0";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(equation);
if (matcher.matches()) {
int a1 = Integer.parseInt(matcher.group(1));
int b1 = Integer.parseInt(matcher.group(2));
int c1 = Integer.parseInt(matcher.group(3));
// System.out.println("a=" + a + "; b=" + b + "; c=" + c);
}
double a = (double) a1; // error message a1 cannot resolve into variable
double b = (double) b1; // error message b1 cannot resolve into variable
double c = (double) c1; // error message c1 cannot resolve into variable
double r1 = 0;
double r2 = 0;
double discriminant = b * b - 4 * a * c;
if (discriminant > 0){
// r = -b / 2 * a;
r1 = (-b + Math.sqrt(discriminant)) / (2 * a);
r2 = (-b - Math.sqrt(discriminant)) / (2 * a);
// System.out.println("Real roots " + r1 + " and " + r2);
}
if (discriminant == 0){
// System.out.println("One root " +r1);
r1 = -b / (2 * a);
r2 = -b / (2 * a);
}
if (discriminant < 0){
// System.out.println(" no real root");
}
String t1 = String.valueOf(r1);
String t2 = String.valueOf(r2);
String t3 ;
t3 = t1+" "+t2;
return t3;
}
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