Question

以下java代码运行良好，并且在“public static void main（String [] args）＆＃39;”下编写时提供了所需的输出。 as：

import com.google.common.collect.ObjectArrays;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import org.apache.commons.lang3.ArrayUtils;
import java.util.regex.*;

public class ArrayReturnSurTime {

public static void main( String[] args )
{ 

     int count1=0;
     int count2=0;
     int total_full_sur=0;
     float word_time= 0.25f;
     float full_sur_time=0.0f;
     float kan_sur_time=0.0f;
     float kun_sur_time=0.0f;

String word="srg]mp";
String modifyword=null;
String after =null;
String rl=null;

char[] c = word.toCharArray();
int m=c.length;
for(int i= 0; i<c.length; i++){
   count1++;

   if(c[i]==']'){count2++;}


 total_full_sur= count1-2*count2;

   full_sur_time=word_time/total_full_sur;
   kan_sur_time = (float) (full_sur_time*0.1);
   kun_sur_time = (float) (full_sur_time*0.9);
}
for(int i=0; i<word.length(); i++){  
char  a =  word.charAt(i);

if( Character.toString(a).equals("]")){
    int j = i-1;
    int k = i+1;
    char b =   word.charAt(j);
    char d =   word.charAt(k);
    modifyword= word.replace( Character.valueOf(b).toString(), ".");
    after=  modifyword.replace( Character.valueOf(d).toString(), ",");

    rl = after;

    String [] sur= word.split("(?!^)");
    String[] tmp_rl= rl.split ("(?!^)");
    float[] rlg_time= new float[sur.length];

         for(int x=0; x<rl.length(); x++){

        switch (tmp_rl[x]) {
            case ".":
                rlg_time[x]=kan_sur_time;
                break;
            case ",":
                rlg_time[x]=kun_sur_time;
                break;
            case "]":
                rlg_time[x]=(float)0.001;
                break;
            default:
                rlg_time[x]=full_sur_time;
                break;
        }

    System.out.print(rlg_time[x]+" ");

 }

}}}
}

它将浮点值打印为：

0.0625 0.0625 0.00625 0.001 0.05625 0.0625

我通过一种方法（ransurtime（String sur36）＆＃39;覆盖代码（需要改变微小代码）来重新编程。返回＆＃39; float [] rlg_time＆＃39;得到相同的输出：

import com.google.common.collect.ObjectArrays;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import org.apache.commons.lang3.ArrayUtils;
import java.util.regex.*;


public class ArrayReturnSurTime {

public static float[] ranisurtime(String sur36) { 


int count1=0;
int count2=0;
int total_full_sur=0;
float word_time= 0.25f;
float full_sur_time=0.0f;
float kan_sur_time=0.0f;
float kun_sur_time=0.0f;

String word=sur36; 
String modifyword=null;String after =null;String rl=null;

char[] c = word.toCharArray();
int m=c.length; 
for(int i= 0; i<c.length; i++){
   count1++; 
   if(c[i]==']'){count2++;} 
 total_full_sur= count1-2*count2; 
   full_sur_time=word_time/total_full_sur;
   kan_sur_time = (float) (full_sur_time*0.1);
   kun_sur_time = (float) (full_sur_time*0.9);
}
for(int p=0; p<word.length(); p++){  
char  a =  word.charAt(p);
if( Character.toString(a).equals("]")) {
    int j = p-1;int k = p+1;char b =   word.charAt(j);char d =   word.charAt(k);
    modifyword= word.replace( Character.valueOf(b).toString(), ".");
    after=  modifyword.replace( Character.valueOf(d).toString(), ",");

    rl = after;

String[] tmp_rl= rl.split ("(?!^)");
float[] rlg_time= new float[tmp_rl.length];

         for(int x=0; x<rl.length(); x++){

        switch (tmp_rl[x]) {
            case ".":
                rlg_time[x]=kan_sur_time;
                break;  
            case ",":
                rlg_time[x]=kun_sur_time;
                break;
            case "]":
                rlg_time[x]=(float)0.001;
                break;
            default:
                rlg_time[x]=full_sur_time;
                break;
        }

         }
         return rlg_time; //This 'rlg_time' is not being returned. Compiler ask return statement although this is the return statement,
}//First closing bracket to be crossed by 'return rlg_time'.
} // Second  closing bracket to be crossed by  'return rlg_time'.
             //return null;//It is the right place at which 'return rlg_time'. statement can do right, but how to manipulate to omit the above two closing brackets reach here.
}  

public static void main( String[] args )
{ 

   String sur36="srg]mp";

float[] A  =    ranisurtime(sur36);

for(int i=0; i<A.length; i++){
  System.out.print(A[i]);
}
}
}

但我失败了。我知道＆＃39;返回rlg_time＆＃39;是不是在正确的地方（它应该移动两个右括号，即＆＃39;}如果要返回它，但是对编程和java不熟悉，我脑海中没有任何东西可以实现它。我需要有助于克服这种情况。

Answer 1

因为你声明了和它没有工作的唯一原因在if语句的正文中定义了rlg_time！什么你试图实现的是返回这个变量的结果方法结束!!!

But, mind you, that the variables are local as per the blocks! Hence, your rlg_time is not returning the proper result.

要解决此问题，请在方法开头将初始化rlg_time与其他变量一起初始化。使用局部变量来使用临时值，例如local。完成reult后，在该局部变量rlg_time的限制（访问点结束）之前指定local此值。现在您可以成功返回rlg_time。顺便说一句，它根本不与Java相关联！

Answer 2

极难阅读您的代码。请正确缩进您的代码。无论如何如果没有其他错误，只需返回类型问题然后将代码更改为： -

import com.google.common.collect.ObjectArrays;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import org.apache.commons.lang3.ArrayUtils;
import java.util.regex.*;


public class ArrayReturnSurTime {

    public static float[] ranisurtime(String sur36) { 


        int count1=0;
        int count2=0;
        int total_full_sur=0;
        float word_time= 0.25f;
        float full_sur_time=0.0f;
        float kan_sur_time=0.0f;
        float kun_sur_time=0.0f;
        float[] rlg_time=null;

        String word=sur36; 
        String modifyword=null;String after =null;String rl=null;

        char[] c = word.toCharArray();
        int m=c.length; 
        for(int i= 0; i<c.length; i++){
            count1++; 
            if(c[i]==']'){count2++;} 
            total_full_sur= count1-2*count2; 
            full_sur_time=word_time/total_full_sur;
            kan_sur_time = (float) (full_sur_time*0.1);
            kun_sur_time = (float) (full_sur_time*0.9);
        }
        for(int p=0; p<word.length(); p++){  
            char  a =  word.charAt(p);
            if( Character.toString(a).equals("]")) {
                int j = p-1;int k = p+1;char b =   word.charAt(j);char d =   word.charAt(k);
                modifyword= word.replace( Character.valueOf(b).toString(), ".");
                after=  modifyword.replace( Character.valueOf(d).toString(), ",");

                rl = after;

                String[] tmp_rl= rl.split ("(?!^)");
                rlg_time= new float[tmp_rl.length]; //changed here

                for(int x=0; x<rl.length(); x++){

                    switch (tmp_rl[x]) {
                    case ".":
                        rlg_time[x]=kan_sur_time;
                        break;  
                    case ",":
                        rlg_time[x]=kun_sur_time;
                        break;
                    case "]":
                        rlg_time[x]=(float)0.001;
                        break;
                    default:
                        rlg_time[x]=full_sur_time;
                        break;
                    }

                }

            }//First closing bracket to be crossed by 'return rlg_time'.
        } // Second  closing bracket to be crossed by  'return rlg_time'.
        return rlg_time;//It is the right place at which 'return rlg_time'. statement can do right, but how to manipulate to reach here.
    }  

    public static void main( String[] args )
    { 

        String sur36="srg]mp";

        float[] A  =    ranisurtime(sur36);

        for(int i=0; i<A.length; i++){
            System.out.print(A[i]);
        }
    }
}

我没有尝试过，但应该可以。您正在初始化FLoat [] array rlg_time in if statment。因此，只要if语句结束，变量就会被释放。将变量声明移到if语句之外。

Answer 3

首先，看一下时间计算的重写：

for(int i= 0; i<word.length(); i++){
    if(word.charAt(i)==']'){count2++;}
}
total_full_sur= word.length() - 2*count2;
full_sur_time=word_time/total_full_sur;
kan_sur_time = (float) (full_sur_time*0.1);
kun_sur_time = (float) (full_sur_time*0.9);

除了消除无用的复杂情况外，请注意您不会在依赖于循环已完成的循环中包含计算。

如果以相同的方式考虑下一个循环，它将解决您的问题。

将退货单转移到正确的位置

3 个答案: