IBM 到 IEEE 浮点转换
java中是否有任何标准方法可以将IBM 370(以字节形式)转换为IEEE格式。任何转换算法都会有所帮助..
我尝试编写java代码..但我不明白我哪里出错了。当我将输入指定为 -2.000000000000000E+02 时,我得到的值为 IEEE 格式的 -140.0。在其他情况下,当我输入 3.140000000000000E+00 时,我得到的值为 IEEE 格式的 3.1712502374909226 对此的任何帮助将不胜感激
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;
}
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;
}
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我可以看到您的问题有几种不同的解决方案:
I can see a couple different solutions to your question:
此 IBM 技术文章包含从 IBM 浮点格式转换为 IEE 浮点的算法。
This IBM Technical Article includes algorithms for converting from IBM floating point formats to IEE floating point.
leftShift() 函数中有一个错误,您应该使用 0x80 而不是 1 进行掩码。这是更正后的函数。
我用 wiki 示例进行了测试 -118.625 如果我理解正确的话,IBM 的偏见会加倍也是 64,所以二进制将是
11000010 01110110 10100000 00000000 00000000 00000000 00000000 00000000
。修复后,程序可以正确生成-118.625。我知道这是一个旧帖子,但我目前也遇到了同样的情况。
There is a bug in the leftShift() function, where you should mask with 0x80 instead of 1. Here is the corrected function.
I tested with the wiki example -118.625 If I understand correctly, the bias for IBM double is also 64, so the binary will be
11000010 01110110 10100000 00000000 00000000 00000000 00000000 00000000
. After the fix, the program can produce -118.625 correctly.I know it is an old post, but I currently ran into the same situation too.