考特长双刊 [英] Cout long double issue

问题描述

所以，我正在从事C ++项目.我有一个long double类型的var，并为其分配了一个值，例如"1.02"

So, I'm working on a C++ project. I have a var of long double type and assigned it a value like "1.02"

然后，我尝试使用cout打印它，结果是:-0

Then, I try to use cout to print it and the result is: -0

我已经尝试使用setprecision，并且发现了所有问题.

I already tried to use setprecision and all I found googling the problem.

对此有什么解决方案?

示例代码:

#include <cstdlib>
#include <iomanip>

using namespace std;

int main(int argc, char** argv)
{

    cout.precision(15);
    long double var = 1.2;
    cout << var << endl;
    return 0;
}

OS:Windows 8.1 64位 编译器:minGW IDE:NetBeans 8.0.2

OS: Windows 8.1 64 bits Compiler: minGW IDE: NetBeans 8.0.2

推荐答案

似乎是编译器有问题.在这里看看: http://mingw.5.n7.nabble.com/Strange-behaviour-of-gcc-4-8-1-with-long-double-td32949.html

It seems to be a problem with compiler. Take a look here: http://mingw.5.n7.nabble.com/Strange-behaviour-of-gcc-4-8-1-with-long-double-td32949.html

使用printf或将变量的值转换为double，然后再传递给cout. (顺便说一句，您确定需要80位精度吗?)

Use printf or convert a value of your variable to double before passing to cout. (BTW are sure you need 80-bit precision?)

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