[英]C++ floor() decreasing value by 1
我編寫了此函數,以將double
精度遞歸舍入到N位數字:
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);
}
}
但是從日志文件來看,在某些情況下floor()確實發生了一些奇怪的事情……
這是良好計算的輸出示例:
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
這是錯誤計算的輸出示例:
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
floor(6770)是否應該返回6770? 為什么返回6769?
因此,首先感謝大家的建議。 順便說一句,“雙倍於字符串->字符串至雙重->底線”的解決方案似乎是唯一給出預期結果的解決方案。 所以我只需要替換:
double floored = floor(sum);
與
double floored = floor(stod(dtos(sum)));
如果有人有更好的解決方案,請發布它。
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