为什么通过cin的int8_t和用户输入显示奇怪的结果 [英] Why does int8_t and user input via cin shows strange result

问题描述

一小段代码让我发疯，但希望你可以阻止我跳出窗外。看这里：

A tiny piece of code drives me crazy but hopefully you can prevent me from jumping out of the window. Look here:

#include <iostream>
#include <cstdint>

int main()
{
    int8_t i = 65;
    int8_t j;

    std::cout << "i = " << i << std::endl; // the 'A' is ok, same as uchar

    std::cout << "Now type in a value for j (use 65 again): " << std::endl;
    std::cin >> j;
    std::cout << "j = " << j << std::endl;

    if (i != j) 
        std::cout << "What is going on here?????" << std::endl;
    else 
        std::cout << "Everything ok." << std::endl;

    return 0;
}

如果我使用 int 而不是 int8_t 一切都好。我需要这个8位无符号整数，而不是更大。顺便说一下。与 unsigned char 它的行为相同 - 当然 - 与 int8_t 一样。

If I use int instead of int8_t everything ok. I need this as 8-bit unsigned integers, not bigger. And btw. with unsigned char it's the same behaviour - of course - as with int8_t.

任何有提示的人？

推荐答案

int8_t 是具有所需特征的整数类型的typedef：纯二进制补码表示，无填充位，大小正好为8位。

int8_t is a typedef for an integer type with the required characteristics: pure 2's-complement representation, no padding bits, size of exactly 8 bits.

对于大多数（可能是全部）编译器，这意味着它将成为 signed char 的typedef。（由于术语有符号整数类型的定义中的怪癖，它不能是普通的 char 的typedef，即使 char 正好签名）。

For most (perhaps all) compilers, that means it's going to be a typedef for signed char.(Because of a quirk in the definition of the term signed integer type, it cannot be a typedef for plain char, even if char happens to be signed).

>> 运算符专门处理字符类型。读取字符会读取单个输入字符，而不是表示十进制中某个整数值的字符序列。因此，如果下一个输入字符是'0'，则读取的值将是字符值'0'，大概是48。

The >> operator treats character types specially. Reading a character reads a single input character, not sequence of characters representing some integer value in decimal. So if the next input character is '0', the value read will be the character value '0', which is probably 48.

由于 typedef 为现有类型创建别名，不是新的不同类型，>> 运算符无法知道您要处理 int8_t 作为整数类型而不是字符类型。

Since a typedef creates an alias for an existing type, not a new distinct type, there's no way for the >> operator to know that you want to treat int8_t as an integer type rather than as a character type.

问题是在大多数实现中没有8位整数类型，它不是字符类型。

The problem is that in most implementations there is no 8-bit integer type that's not a character type.

唯一的解决方法是读入 int 变量，然后转换为 int8_t （如果需要，可以进行范围检查）。

The only workaround is to read into an int variable and then convert to int8_t (with range checks if you need them).

顺便说一下， int8_t 是签名类型;相应的无符号类型是 uint8_t ，其范围为0..255。

Incidentally, int8_t is a signed type; the corresponding unsigned type is uint8_t, which has a range of 0..255.

（还有一个考虑因素：如果 CHAR_BIT> 8 ，这是标准允许的，那么 int8_t 和 uint8_t 将被定义。）

(One more consideration: if CHAR_BIT > 8, which is permitted by the standard, then neither int8_t nor uint8_t will be defined at all.)

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