问题描述
您好我都按照基础10浮点值： 0.625 。我需要转换以10为二进制格式该值是： 0.101 。算法我发现如下。它的工作原理，但我不理解为什么这个工程。可能有人请解释低于工程为什么code？据我所知，小数点后的数字计算的方式，使得1/2 ^ n，其中n是从小数点计数。谢谢你。
Hi I have following float point value in base 10: 0.625. I need to convert this value in base 10 to binary format which is: 0.101.Algorithm I found is below. It works, but I am not understanding why this works. Could someone please explain to be why the code below works? I am aware that numbers after the decimal point is computed in the manner such that 1/2^n where n is count from the decimal point. Thanks.
要澄清一下，我需要知道的数学公式背后的原因。不步进通过code。
To clarify, I need to know the reasoning behind the mathematical formula. Not stepping through the code.
private static String floatToBinaryString( double n ) {
String val = "0.";
while ( n > 0 ) {
double r = n * 2;
if( r >= 1 ) {
val += "1";
n = r - 1;
}else{
val += "0";
n = r;
}
}
return val;
}
推荐答案
您乘以2馏分并使用个位数字为二进制值，直到馏分等于零。下面的例子。
You multiply the fraction by 2 and use the ones place digit as the binary values until the fraction is equal to zero. Example below.
这是标准的公式使用转换的0.625有：
This is the standard formula for conversion using the 0.625 you have:
1) Multiply fraction by 2 => 0.625 * 2 = 1.25
The digit to the left of the decimal point is the first binary value, 0.1 so far
2) Ignore the ones-place digit and you have 0.25 which is still larger than zero.
Multiply the fraction by 2 => 0.25 * 2 = 0.50
The digit to the left of the decimal point is the next binary value, 0.10 so far
3) Ignore the ones-place digit and you have 0.50 which is less than zero.
Multiply the fraction by 2 => 0.5 * 2 = 1.00
The digit to the left of the decimal point is the next binary value, 0.101 so far
4) Ignore the ones-place digit and you have 0.00 which is equal to zero.
Conversion complete!
private static String floatToBinaryString( double n ) {
String val = "0."; // Setting up string for result
while ( n > 0 ) { // While the fraction is greater than zero (not equal or less than zero)
double r = n * 2; // Multiply current fraction (n) by 2
if( r >= 1 ) { // If the ones-place digit >= 1
val += "1"; // Concat a "1" to the end of the result string (val)
n = r - 1; // Remove the 1 from the current fraction (n)
}else{ // If the ones-place digit == 0
val += "0"; // Concat a "0" to the end of the result string (val)
n = r; // Set the current fraction (n) to the new fraction
}
}
return val; // return the string result with all appended binary values
}
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