递归的流程,java崩溃的原因
- Java code
import java.util.*;public class CrashJava{ public String toString() { return "CrashJava Address: " + this + "\n"; } public static void main(String[] args) { Vector v = new Vector(); for(int i = 0; i < 10; i++) { v.addElement(new CrashJava()); } System.out.println(v); }}[解决办法]
public String toString()
{
return "CrashJava Address: " + this + "\n";/
}
等同于
public String toString()
{
return "CrashJava Address: " + this.toString() + "\n";
}