/**
* Produces a string form of an unsigned given the value and the numerical base to convert it to.
* This class method can be used by anyone anytime. If base is 16, result is same as for
* formatNumber(). If base is 10, will produce string version of unsigned value. E.g. 0xffffffff
* will produce "4294967295" instead of "-1".
*
* @param value the number to be converted
* @param base the numerical base to use (currently 10 or 16)
* @return a String equivalent of the value rendered appropriately.
*/
public static String formatUnsignedInteger(int value, int base) {
if (base == NumberDisplayBaseChooser.HEXADECIMAL) {
return Binary.intToHexString(value);
} else {
return Binary.unsignedIntToIntString(value);
}
}
/**
* Produces a string form of a double given the value and the numerical base to convert it to.
* There is an instance method that uses the internally stored base. This class method can be used
* by anyone anytime.
*
* @param value the number to be converted
* @param base the numerical base to use (currently 10 or 16)
* @return a String equivalent of the value rendered appropriately.
*/
public static String formatNumber(double value, int base) {
if (base == NumberDisplayBaseChooser.HEXADECIMAL) {
long lguy = Double.doubleToLongBits(value);
return Binary.intToHexString(Binary.highOrderLongToInt(lguy))
+ Binary.intToHexString(Binary.lowOrderLongToInt(lguy)).substring(2);
} else {
return Double.toString(value);
}
}
/**
* Produces a string form of a double given a long containing the 64 bit pattern and the numerical
* base to use (10 or 16). If the base is 16, the string will be built from the 64 bits. If the
* base is 10, the long bits will be converted to double and the string constructed from that.
* Seems an odd distinction to make, except that contents of floating point registers are stored
* internally as int bits. If the int bits represent a NaN value (of which there are many!),
* converting them to double then calling formatNumber(double, int) above, causes the double value
* to become the canonical NaN value. It does not preserve the bit pattern! Then converting it to
* hex string yields the canonical NaN. Not an issue if display base is 10 since result string
* will be NaN no matter what the internal NaN value is.
*
* @param value the long bits to be converted to string of corresponding double.
* @param base the numerical base to use (currently 10 or 16)
* @return a String equivalent of the value rendered appropriately.
*/
public static String formatDoubleNumber(long value, int base) {
if (base == NumberDisplayBaseChooser.HEXADECIMAL) {
return Binary.longToHexString(value);
} else {
return Double.toString(Double.longBitsToDouble(value));
}
}
/**
* Produces a string form of a float given an integer containing the 32 bit pattern and the
* numerical base to use (10 or 16). If the base is 16, the string will be built from the 32 bits.
* If the base is 10, the int bits will be converted to float and the string constructed from
* that. Seems an odd distinction to make, except that contents of floating point registers are
* stored internally as int bits. If the int bits represent a NaN value (of which there are
* many!), converting them to float then calling formatNumber(float, int) above, causes the float
* value to become the canonical NaN value 0x7fc00000. It does not preserve the bit pattern! Then
* converting it to hex string yields the canonical NaN. Not an issue if display base is 10 since
* result string will be NaN no matter what the internal NaN value is.
*
* @param value the int bits to be converted to string of corresponding float.
* @param base the numerical base to use (currently 10 or 16)
* @return a String equivalent of the value rendered appropriately.
*/
public static String formatFloatNumber(int value, int base) {
if (base == NumberDisplayBaseChooser.HEXADECIMAL) {
return Binary.intToHexString(value);
} else {
return Float.toString(Float.intBitsToFloat(value));
}
}
/**
* Produces a string form of a number given the value. There is also an class (static method) that
* uses a specified base.
*
* @param value the number to be converted
* @return a String equivalent of the value rendered appropriately.
*/
public String formatNumber(int value) {
if (base == NumberDisplayBaseChooser.HEXADECIMAL) {
return Binary.intToHexString(value);
} else {
return new Integer(value).toString();
}
}
/**
* Produces a string form of an integer given the value and the numerical base to convert it to.
* There is an instance method that uses the internally stored base. This class method can be used
* by anyone anytime.
*
* @param value the number to be converted
* @param base the numerical base to use (currently 10 or 16)
* @return a String equivalent of the value rendered appropriately.
*/
public static String formatNumber(int value, int base) {
String result;
switch (base) {
case HEXADECIMAL:
result = Binary.intToHexString(value);
break;
case DECIMAL:
result = Integer.toString(value);
break;
case ASCII:
result = Binary.intToAscii(value);
break;
default:
result = Integer.toString(value);
}
return result;
// if (base == NumberDisplayBaseChooser.HEXADECIMAL) {
// return Binary.intToHexString(value);
// }
// else {
// return Integer.toString(value);
// }
}
