从boost cpp_rative到int的转换错误

Question

boost cpp_rative似乎在分子和分母有多个数字时错误地转换为int。

#include <iostream>
#include <boost/multiprecision/cpp_int.hpp>
using namespace boost::multiprecision;
using namespace std;
int main() {
cpp_rational a("4561231231235/123123123123");
std::cout << "bad convert: " << a << ' '  << 
float(a) << ' ' << int(a) << ' ' << 
a.convert_to<int>() << endl;
a = (cpp_rational)"456/123";
std::cout << "good convert: " << a << ' '  <<
float(a) << ' ' << int(a) << ' ' << 
a.convert_to<int>() << endl;
}

输出为：

bad convert: 651604461605/17589017589 37.0461 -3 -3
good convert: 152/41 3.70732 3 3

此外，将cpp_rative转换为cpp_int的尝试也无法编译，例如，使用

cpp_int b = static_cast<cpp_int> (a);
cpp_int b = a.convert_to<cpp_int>();

我想做的是进行除法和四舍五入，即使接近整数也不会出错。

帮助？谢谢

Answer 1

文档说

也允许转换：

d = a; // OK, widening conversion.
d = a * b;  // OK, can convert from an expression template too.

然而，本质上有损的转换要么被明确宣布，要么被完全禁止：

d = 3.14;  // Error implicit conversion from float not allowed.
d = static_cast<mp::int512_t>(3.14);  // OK explicit construction is allowed

---

所以，您看到的是有损转换。证明如果您没有明确指定，转换将被禁止：

cpp_rational a("4561231231235/123123123123");
int i   = a;        // error: cannot convert `cpp_rational` to `int` in initialization
float f = a;        // error: cannot convert `cpp_rational` to `float` in initialization
long double ld = a; // error: cannot convert `cpp_rational` to `long double` in initialization
long long   ll = a; // error: cannot convert `cpp_rational` to `long long int` in initialization

所以，你对所有显式强制转换所做的就是告诉编译器"闭嘴，我知道我在做什么"。所以你为什么没有得到警告。

现在，如何解决问题：将除法保持在多精度域：

cpp_int v = numerator(a)/denominator(a);

您的样本已修复：在Coliru上直播

#include <boost/multiprecision/cpp_int.hpp>
#include <iostream>
using namespace boost::multiprecision;
int main() {
cpp_rational a("4561231231235/123123123123");
cpp_int v = numerator(a)/denominator(a);
std::cout << "convert: " << v.convert_to<int>() << "n";
a = cpp_rational("456/123");
v = numerator(a)/denominator(a);
std::cout << "convert: " << v.convert_to<int>() << "n";
}