CODEWITHSUNDEEP
javamd5decodeencode
zzy
zzy

Reputation: 1793

can ISO-8859-1 encode MD5 bytes correctly ?

1.There is some data in String:

String data = "some......";

2.And use MD5 covert it into bytes:

byte [] result = MD5.toMD5(data);

3.Now I encode it into String:

String encodeString = new String(result,"ISO-8895-1");

4.And then decode it to bytes:

byte [] decodeBytes = encodeString.getBytes("ISO-8859-1");

Here my question is : Will the decodeBytes be equal to result ?

My confusion is whether there will be Zero in result and will it cause truncation in Step3 ?

If there is any problem to let decodeBytes equal to result, and if I limit the data type of String in Step1 such as just allow Letters and Numbers , will the problem can be avoid?

Upvotes: 1

Views: 423

Answers (1)

laune
laune

Reputation: 31300

There is no reason why a byte value should not be decoded to a character if ISO-8859-1, which is a 8 bit character code. Although 65 code points (for control characters) are not included, the String methods handle these as if the conrols defined in ISO/IEC 6429 were part of that character set.

Round tripping byte values 0 to 255 works perfectly, also with byte[].

byte[] bs = new byte[256];
String encode() throws Exception {
    return new String( bs, "ISO-8859-1" );
}
byte[] decode( String s ) throws Exception{
    return s.getBytes( "ISO-8859-1" );
}
 void set(){
    for( int i = 0; i < bs.length; ++i ){
        bs[i] = (byte)i;
    }
}
boolean cmp( byte[] x ){
    for( int i = 0; i < bs.length; ++i ){
        if( bs[i] != x[i] ){
            System.out.println( i + ": " + bs[i] + " != " + x[i] );
            return false;
         }
    }
    return true;
}
void round() throws Exception{
    String s = encode();
    if( s.length() != 256 ) throw new IllegalStateException();
        byte[] res = decode( s );
        if( ! cmp( res ) ) System.out.println( "false" );
    }
}

Upvotes: 2

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