我一直发现 iomanip 令人困惑和反直觉。我需要帮助。
快速互联网搜索发现(https://www.vedantu.com/maths/precision)“因此,我们将精度视为十进制数中小数点后有效数字的最大数量”(重点是我的)。这也符合我的理解。但是我写了一个测试程序并且:
stm << std::setprecision(3) << 5.12345678;
std::cout << "5.12345678: " << stm.str() << std::endl;
stm.str("");
stm << std::setprecision(3) << 25.12345678;
std::cout << "25.12345678: " << stm.str() << std::endl;
stm.str("");
stm << std::setprecision(3) << 5.1;
std::cout << "5.1: " << stm.str() << std::endl;
stm.str("");
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输出:
5.12345678: 5.12
25.12345678: 25.1
5.1: 5.1
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如果精度为 3,则输出应为:
5.12345678: 5.123
25.12345678: 25.123
5.1: 5.1
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显然,C++ 标准对与浮点数相关的“精度”的含义有不同的解释。
如果我做:
stm.setf(std::ios::fixed, std::ios::floatfield);
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然后前两个值的格式正确,但最后一个值显示为5.100.
如何在没有填充的情况下设置精度?
您可以尝试使用此解决方法:
decltype(std::setprecision(1)) setp(double number, int p) {
int e = static_cast<int>(std::abs(number));
e = e != 0? static_cast<int>(std::log10(e)) + 1 + p : p;
while(number != 0.0 && static_cast<int>(number*=10) == 0 && e > 1)
e--; // for numbers like 0.001: those zeros are not treated as digits by setprecision.
return std::setprecision(e);
}
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进而:
auto v = 5.12345678;
stm << setp(v, 3) << v;
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另一个更详细和优雅的解决方案是创建一个像这样的结构:
struct __setp {
double number;
bool fixed = false;
int prec;
};
std::ostream& operator<<(std::ostream& os, const __setp& obj)
{
if(obj.fixed)
os << std::fixed;
else os << std::defaultfloat;
os.precision(obj.prec);
os << obj.number; // comment this if you do not want to print immediately the number
return os;
}
__setp setp(double number, int p) {
__setp setter;
setter.number = number;
int e = static_cast<int>(std::abs(number));
e = e != 0? static_cast<int>(std::log10(e)) + 1 + p : p;
while(number != 0.0 && static_cast<int>(number*=10) == 0)
e--; // for numbers like 0.001: those zeros are not treated as digits by setprecision.
if(e <= 0) {
setter.fixed = true;
setter.prec = 1;
} else
setter.prec = e;
return setter;
}
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像这样使用它:
auto v = 5.12345678;
stm << setp(v, 3);
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