使用 IOS 成员进行格式化

发布于 2024-11-28 19:46:58 字数 531 浏览 0 评论 0原文

我有以下代码，只需看一下它，

#include<iostream>
#include<conio.h>
#include<string>
using namespace std;

int main()
{
    float a=56;
    cout.setf(ios::hex);

    cout<<"\nyou have entered "<<a;/*this statement must output a value in hexadecimal*/
    _getch();
    cout.unsetf(ios::hex);
    cout<<"\n modified value"<<a; /*& it should give me an answer 56*/

    _getch();
    return 0;
}

但第一个注释语句对我不起作用，它还打印出 56。我是否做错了，或者其他什么？
（我正在使用 Visual C++ 编译器）。

原文

I have the following code, just have a look at it

#include<iostream>
#include<conio.h>
#include<string>
using namespace std;

int main()
{
    float a=56;
    cout.setf(ios::hex);

    cout<<"\nyou have entered "<<a;/*this statement must output a value in hexadecimal*/
    _getch();
    cout.unsetf(ios::hex);
    cout<<"\n modified value"<<a; /*& it should give me an answer 56*/

    _getch();
    return 0;
}

but the first commented statement is not working for me, it also prints out 56. Am I doing a mistake, or anything else?
(I am using a visual c++,compiler).

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流殇 2024-12-05 19:46:58

您必须使用 setf 的两个参数版本，因为格式标志的基本设置不是单个位，而是使用多个位：

std::cout.setf( std::ios_base::hex, std::ios_base::basefield );

setf 的两个参数版本code> 确保需要清除的basefield 位确实被清除。

同样，您不能“取消设置”hex，因为它不是单个位，您必须设置不同的基数：

std::cout.setf( std::ios_base::dec, std::ios_base::basefield );

最重要的是：请注意，在当前标准中，的十六进制格式>ostream 仅适用于整数，不适用于浮点类型。您将需要使用或转换为整数才能查看十六进制输出。

为了避免所有疑问，此代码示例按预期“工作”：

#include <iostream>
#include <ostream>

int main()
{
    int a = 56;
    std::cout.setf( std::ios_base::hex, std::ios_base::basefield );
    std::cout << "Hex: " << a << '\n';
    std::cout.setf( std::ios_base::dec, std::ios_base::basefield );
    std::cout << "Dec: " << a << '\n';
    return 0;
}

输出：

Hex: 38
Dec: 56

You have to use the two argument version of setf because the base settings of the format flags isn't a single bit, it uses a number of bits:

std::cout.setf( std::ios_base::hex, std::ios_base::basefield );

The two parameter version of setf ensures that basefield bits that need to be cleared are actually cleared.

Similarly, you can't "unset" hex because it's not a single bit, you have to set a different base:

std::cout.setf( std::ios_base::dec, std::ios_base::basefield );

Most importantly: note that in the current standard, hexadecimal formatting for ostream only applies to integers, not floating point types. You will need to use or cast to an integer to see a hexadecimal output.

For the avoidance of all doubt this code sample "works" as expected:

#include <iostream>
#include <ostream>

int main()
{
    int a = 56;
    std::cout.setf( std::ios_base::hex, std::ios_base::basefield );
    std::cout << "Hex: " << a << '\n';
    std::cout.setf( std::ios_base::dec, std::ios_base::basefield );
    std::cout << "Dec: " << a << '\n';
    return 0;
}

Output: