Question

当我尝试减去最高值时，我的c ++代码返回扩展表示法。

    vector<std::string> smarkdat
    vector<std::string> markdat;
    vector<double> doubleMdat;
    vector<double> sdoubleMdat;
    markdat = getmarketbuyData();
    smarkdat = getmarketsellData();
    transform(smarkdat.begin(), smarkdat.end(), back_inserter(sdoubleMdat), [](string const& val) {return stod(val);});
    auto sbiggestMark = std::max_element(std::begin(sdoubleMdat), std::end(sdoubleMdat));
    transform(markdat.begin(), markdat.end(), back_inserter(doubleMdat), [](string const& val) {return stod(val);});
    auto biggestMark = std::max_element(std::begin(doubleMdat), std::end(doubleMdat));
    std::cout << "The highest seller is " << *sbiggestMark << " at position " << std::distance(std::begin(sdoubleMdat), sbiggestMark) << std::endl;
    std::vector<double> difference(sdoubleMdat.size(), 0.0);

    for ( std::vector<double>::size_type i = 0; i < sdoubleMdat.size(); i++ )
        {
             difference[i] = std::abs( sbiggestMark[i] - biggestMark[i] );
             cout << difference[i] << endl;
        }

输出：

8.636e-05

矢量数据是字符串0.000000200转换为double。

如何返回2个双倍的常规减去值而不是非常明显的错误返回值？

Answer 1

如果问题是您想要在没有科学记数法的情况下设置输出，请使用std::setprecision中的std::fixed和<iomanip>：

#include <iostream>
#include <iomanip>
#include <cmath>

using namespace std;

int main()
{
    double x = fabs(sqrt(2.0) - sqrt(2.0001));
    cout << x << "\n"; 
    cout << setprecision(10) << fixed << x << '\n';
}

实例：http://ideone.com/B3IfDF

减去双向量返回扩展表示法

1 个答案: