在C ++中重置输出标志 [英] Resetting output flags in C++

问题描述

我打算在使用resetiosflags函数结束的行上将所有输出标志重置为默认值.当我尝试以这种方式进行操作时，它会提供错误的输出，这与我的预期相反.

I'm intending to reset all output flags to default on the lines where I end using the resetiosflags function. It provides erroneous output when I attempt to do it in this manner, contrary to my expectations.

#include <iostream>
#include <iomanip>
using namespace std;
int
main()
{
    bool first;
    int second;
    long third;
    float fourth;
    float fifth;
    double sixth;

    cout << "Enter bool, int, long, float, float, and double values: ";
    cin >> first >> second >> third >> fourth >> fifth >> sixth;
    cout << endl;

// ***** Solution starts here ****
    cout << first << " " << boolalpha << first  << endl << resetiosflags;
    cout << second << " " << showbase << hex << second << " " << oct << second << endl << resetiosflags;
    cout << third << endl;
    cout << showpos << setprecision(4) << showpoint << right << fourth << endl << resetiosflags;
    cout << scientific << fourth << endl << resetiosflags;
    cout << setprecision(7) << left << fifth << endl << resetiosflags;
    cout << fixed << setprecision(3) << fifth << endl << resetiosflags;
    cout << third << endl;
    cout << fixed << setprecision(2) << fourth << endl << resetiosflags;
    cout << fixed << setprecision(0) << sixth << endl << resetiosflags;
    cout << fixed << setprecision(8) << fourth << endl << resetiosflags;
    cout << setprecision(6) << sixth << endl << resetiosflags;
// ***** Solution ends here ****

    cin.get();
    return 0;
}

我已知的替代方法是通过重新声明它们来分别对其进行标记，但这似乎是多余的.

My known alternative is de-flagging them individually by restating them, but that seems superfluous.

推荐答案

/*unspecified*/ resetiosflags( std::ios_base::fmtflags mask );

std::resetiosflags()是打算在诸如out << resetiosfloags( flags )之类的表达式中使用的操纵器.大概您正在做的是传入一个函数指针，该指针由std::operator<<的重载选择，该重载采用布尔值并输出1.

std::resetiosflags() is a manipulator intended to be used in an expression such as out << resetiosfloags( flags ). Presumably what you're doing is passing in a function pointer, which gets selected by the overload of std::operator<< that takes a boolean and prints 1.

但是std::resetiosflags()将格式标志用作无法操纵精度的参数.但是，std::ios_base::boolalpha可以:

But std::resetiosflags() takes a format flags as a parameter which the precision can't be manipulated with. std::ios_base::boolalpha can, however:

std::cout << ... << std::resetiosflags(std::ios_base::boolalpha);

还有std::noboolalpha

std::cout << ... << std::noboolalpha;

但是，如果您需要将精度重置为默认值，则可以为此创建自己的操纵器.您还可以使用提升IO状态保护程序.

But if you need to reset the precision to its default you can just create your own manipulator for that. You can also use Boost IO State Saver.

这篇关于在C ++中重置输出标志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IT屋！