1字节整数数据类型 [英] 1 byte integer data type

问题描述

我写了以下代码：

  #include< iostream& 
 #include< iomanip> 
 #include< stdint.h> 
 
 using namespace std; 
 
 int main（）
 {
 uint8_t c; 
 
 cin>> hex>> C; 
 cout<< dec<< C; 
 
 return 0; 
} 
  

但是当我输入 c — hex for 12—输出也是 c 。我期待着12.后来我了解到：

uint8_t 通常是typedef unsigned char 。因此，它实际上是以ASCII 0x63读取 c 。

整数在执行I / O时不作为整数，而不是作为char？

解决方案

不是我知道的。 >

您可以使用更宽的整数类型来执行I / O，并根据需要使用范围检查和转换。

I wrote the following code:

 #include <iostream>
 #include <iomanip>
 #include <stdint.h>

 using namespace std;

 int main()
 {
     uint8_t c;

     cin  >> hex >> c;
     cout << dec << c;

     return 0;
 }

But when I input c—hex for 12—the output is also c. I was expecting 12. Later I learned that:

uint8_t is usually a typedef for unsigned char. So it's actually reading c as ASCII 0x63.

Is there a 1 byte integer which behaves as an integer while doing I/O and not as char?

解决方案

Not that I know of.

You could do the I/O using a wider integer type, and use range checking and casting as appropriate.

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