Cout 长双发行

问题描述

所以，我正在做一个 C++ 项目.我有一个 long double 类型的 var 并为其分配了一个类似1.02"的值

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，我尝试用cout打印出来，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 并且我在谷歌搜索中发现了所有问题.

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;
}

操作系统:Windows 8.1 64 位编译器:minGWIDE:NetBeans 8.0.2

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

推荐答案

似乎是编译器的问题.看看这里: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 或在传递给 cout 之前将变量的值转换为 double.(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?)

这篇关于Cout 长双发行的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！