Java use String.valueOf(char[]) but shows garbled

Question

char[] string = new char[16];
for(int i = 15; i >= 0; i--,input>>=1){
    string[i] = (char)(input&1+48);//ascii '0'=48
}
return String.valueOf(string);

use this way, will get garbled output as follow:

Input:

Output:

(will show binary square)

What is `input`? Please provide a [mcve], with special emphasis on COMPLETE. — Jim Garrison, Sep 13 '17 at 06:35
To the user who dup-hammered this, I respectfully submit that the question is too unclear to be a dup of anything. Closing it as a dup enshrines it as legitimate, when it probably needs to be closed as off-topic or unclear. — Jim Garrison, Sep 13 '17 at 06:41
@JimGarrison You're right...I saw the array and garbled output and jumped the gun. — Tim Biegeleisen, Sep 13 '17 at 06:53
I upvoted your question, please clean it up and make it coherent. — Tim Biegeleisen, Sep 13 '17 at 06:57

score 1 · Answer 1 · answered Sep 13 '17 at 06:37

Analysis about this Error:

char[] string2 = {(char)(0&1+48),(char)48,'0','0','0','0','0','0','0','0','0','0','0','0','0','0'};//Test
char[] string2 = {(char)(((int)0&1)+48),(char)48,'0','0','0','0','0','0','0','0','0','0','0','0','0','0'};//Test
System.out.println(string2);//Test
System.out.println(String.valueOf(string2));//Test

When I write "(char)(((int)0&1)+48)" the IDE told me cast to int is redundant

The Right Code:

char[] string = new char[16];
for(int i = 15; i >= 0; i--,input>>=1){
    string[i] = (char)((int)(input&1)+48);//ascii '0'=48
}
return String.valueOf(string);

Second Way:

StringBuilder sb = new StringBuilder("");
for(int i = 15; i >= 0; i--,input>>=1){
    sb.insert(0,String.valueOf(input&1));
}
return sb.toString();

Java use String.valueOf(char[]) but shows garbled

1 Answers1