IBM到IEEE浮点转换

问题描述

在java中有没有标准的方法来将IBM 370（以字节的形式）转换为IEEE格式。转换的任何算法都会有帮助..

我试图写一个Java代码..但我不明白我在哪里出错。当我给输入-2.000000000000000E + 02，我得到的值为-140.0在IEEE格式。并在其他情况下，当我给输入为3.140000000000000E + 00我得到的价值为3.1712502374909226 IEEE格式任何帮助，将不胜感激

  private void conversion（）{
 byte [] buffer = //要读取的字节（8字节）; 
 int sign =（buffer [0]& 0x80）; 
 //提取指数。 
 int exp =（（buffer [0]& 0x7f） -  64）* 4  -  1; 
 //标准化尾数。 （（缓冲区[1]& 0x80）== 0）{$ b $（$ i = 0; i< 4; i ++） b buffer = leftShift（buffer）; 
 exp = exp  -  1; 
 
 
 
 //把符号和尾数换成8个字节的数字
 buffer = rightShift（buffer）; //为更长的指数腾出空间
 buffer = rightShift（buffer）; 
 buffer = rightShift（buffer）; 
 exp = exp + 1023; //过量的1023格式
 int temp = exp& 0x0f; //低4位进入B（1）
 buffer [1] =（byte）（（buffer [1]& 0xf）|（temp * 16））; 
 buffer [0] =（byte）（sign |（exp / 16）& 0x7f））; 
 
 
 private byte [] rightShift（byte [] buf）{
 int newCarry = 0; 
 int oldCarry = 0; 
 for（int i = 1; i  newCarry = buf [i]& 1; 
 buf [i] =（byte）（（buf [i]& 0xFE）/ 2 +（oldCarry！= 0？0x80：0））; 
 oldCarry = newCarry; 
} 
 return buf; 
 
 
 private byte [] leftShift（byte [] buf）{
 int newCarry = 0; 
 int oldCarry = 0; 
 for（int i = buf.length-1; i> 0; i--）{
 newCarry = buf [i]& 1; 
 buf [i] =（byte）（（buf [i]& 0x7F）* 2 +（oldCarry！= 0？1：0））; 
 oldCarry = newCarry; 
} 
 return buf; 
 
  

解决方案

包含从IBM浮点格式转换为IEE浮点的算法。

Is there any standard method in java to convert IBM 370(in the form of bytes) to IEEE format.?Any algorithm for the conversion would help..

I tried writing a java code..But i fail to understand where do i go wrong. When i give the input as -2.000000000000000E+02, i'm getting the value as -140.0 in IEEE format. and in othercase when i give the input as 3.140000000000000E+00 i'm getting the value as 3.1712502374909226 in IEEE format Any help on this would be highly appreciated

private void conversion() {
    byte[] buffer = //bytes to be read(8 bytes);
    int sign = (buffer[0] & 0x80);
    // Extract exponent.
    int exp = ((buffer[0] & 0x7f) - 64) * 4 - 1;
    //Normalize the mantissa.
    for (int i = 0; i < 4; i++) {//since 4 bits per hex digit
        if ((buffer[1] & 0x80) == 0) {
            buffer = leftShift(buffer);
            exp = exp - 1;
        }
    }

    // Put sign and mantissa back in 8-byte number
    buffer = rightShift(buffer);// make room for longer exponent
    buffer = rightShift(buffer);
    buffer = rightShift(buffer);
    exp = exp + 1023;//Excess 1023 format
    int temp = exp & 0x0f;//Low 4 bits go into B(1)
    buffer[1]= (byte)((buffer[1]&0xf) | (temp *16));
    buffer[0]= (byte)(sign | ((exp/16) & 0x7f));
    }

     private byte[] rightShift(byte[] buf) {
    int newCarry = 0;
    int oldCarry = 0;
    for(int i = 1; i<buf.length; i++) {
        newCarry = buf[i] & 1;
        buf[i] = (byte)((buf[i] & 0xFE)/2 + (oldCarry != 0 ? 0x80 : 0));
        oldCarry = newCarry;
    }
    return buf;
}

private byte[] leftShift(byte[] buf) {
    int newCarry = 0;
    int oldCarry = 0;
    for(int i = buf.length-1; i>0; i--) {
        newCarry = buf[i] & 1;
        buf[i] = (byte)((buf[i] & 0x7F)*2 + (oldCarry != 0 ? 1 : 0));
        oldCarry = newCarry;
    }
    return buf;
}

解决方案

This IBM Technical Article includes algorithms for converting from IBM floating point formats to IEE floating point.

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