无符号字符的标准输入 [英] Standard Input for Unsigned Character

问题描述

我试图通过程序发送无符号字符，我想能够通过标准输入（即std :: cin）获取数字。例如，当我输入2我想它发送☻（unsigned char 2）。当我使用代码：

  std :: cout< 输入值：; 
 {
 unsigned char d; 
 unsigned char e = 2; 
 std :: cin>> d; 
 WriteFile（file，& d，1，& written，NULL）; 
 std :: cout<< d =<< d<< \\\
; 
 std :: cout<< e =< e; 
} 
  

我得到

 输入值：2 
d = 2 
e =☻
  

任何人都可以告诉我为什么d被解释不正确为unsigned char 50而e被正确解释为unsigned char 2？

当然，在您解释之后，您可以解释如何获取用户输入并将其转换，以便我发送2而不是2。

解决方案

因为 std :: cin>> d; 默认读取 char 类型，因此输入 2 2 （ASCII代码为50），而不是由ASCII代码2表示的字符。这是正常的行为，否则尝试从 cin c>



另一方面，在 unsigned char e = 2; 你显式地为变量赋值（ 2 ），所以编译器盲目地赋值给 e

您可能想要这样：

  #include< iostream> ; 
 #include< string> 
 using namespace std; 
 
 int main（int argc，char * argv []）
 {
 string myString; 
 cin>> myString; 
 char c = atoi（myString.c_str（））; 
 cout<< c<< endl; 
} 
  

I am trying to send unsigned characters through a program, and I would like to be able to get the numbers through standard input (ie std::cin). For example when I type 2 I would like it send ☻ (unsigned char 2). when I use the code:

 std::cout << "Enter values: ";
            {
                unsigned char d;
                unsigned char e = 2;
                std::cin >> d;
                WriteFile(file, &d, 1, &written, NULL);
                std::cout << "d= " << d << "\n"; 
                std::cout << "e= " << e;
            }

I get

 Enter values: 2
 d=2
 e=☻

Can anyone tell me why d is being interpreted Incorrectly as unsigned char 50 while e is being interpreted correctly as unsigned char 2?

And of course after your explanation can You explain how to get User input and convert it so that I send 2 rather than '2'.

解决方案

Because std::cin >> d; reads by default a char type, so the input 2 translates into the character 2 (with ASCII code 50) and not the character represented by the ASCII code 2. This is a normal behaviour, otherwise trying to read numbers from cin will end up being a mess.

On the other hand, in unsigned char e = 2; you explicitly assign a value (2) to the variable, so the compiler blindly assigns it to e.

You probably want this:

#include <iostream>
#include <string>
using namespace std;

int main(int argc, char *argv[])
{
    string myString;
    cin >> myString;
    char c = atoi(myString.c_str());
    cout << c << endl;
}

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