0
值我寫了這個函數遞歸一輪double
到N位數:C++地板()減少1
double RoundDouble(double value, unsigned int digits)
{
if (value == 0.0)
return value;
string num = dtos(value);
size_t found = num.find(".");
string dec = "";
if (found != string::npos)
dec = num.substr(found + 1);
else
return value;
if (dec.length() <= digits)
{
LogToFile("C:\\test.txt", "RETURN: " + dtos(value) + "\n\n\n");
return value;
}
else
{
double p10 = pow(10, (dec.length() - 1));
LogToFile("C:\\test.txt", "VALUE: " + dtos(value) + "\n");
double mul = value * p10;
LogToFile("C:\\test.txt", "MUL: " + dtos(mul) + "\n");
double sum = mul + 0.5;
LogToFile("C:\\test.txt", "SUM: " + dtos(sum) + "\n");
double floored = floor(sum);
LogToFile("C:\\test.txt", "FLOORED: " + dtos(floored) + "\n");
double div = floored/p10;
LogToFile("C:\\test.txt", "DIV: " + dtos(div) + "\n-------\n");
return RoundDouble(div, digits);
}
}
但是從日誌文件中,一些真正奇怪的是,在某些情況下與地面發生() ...
這裏有良好的計算的輸出示例:
VALUE: 2.0108
MUL: 2010.8
SUM: 2011.3
FLOORED: 2011
DIV: 2.011
-------
VALUE: 2.011
MUL: 201.1
SUM: 201.6
FLOORED: 201
DIV: 2.01
-------
RETURN: 2.01
而這裏的壞計算的輸出示例:
VALUE: 67.6946
MUL: 67694.6
SUM: 67695.1
FLOORED: 67695
DIV: 67.695
-------
VALUE: 67.695
MUL: 6769.5
SUM: 6770
FLOORED: 6769 <= PROBLEM HERE
DIV: 67.69
-------
RETURN: 67.69
不是地板(6770)應該返回6770嗎?爲什麼它返回6769?
因爲它實際上不是6770,而是非常接近它。 [每個計算機科學家應該知道什麼關於浮點運算](http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html) –
「是不是應該返回(6770) 6770?」對於6770的小值,這是不正確的。 – user2079303
嘗試在地板上添加一個epsilon值,聽起來這個值真的是6769,小數點接近6770. –