uint8_t iostream行为 [英] uint8_t iostream behavior

问题描述

摘要：我期待的代码：cout< uint8_t（0）;打印0，但它不打印任何东西。

Abstract: I was expecting the code: cout << uint8_t(0); to print "0", but it doesn't print anything.

长版本：当我尝试将uint8_t对象流化到cout时，这是预期的行为吗？可能是uint8_t是一些char类型的别名？请参阅代码示例中的编译器/系统注释。

Long version: When I try to stream uint8_t objects to cout, I get strange characters with gcc. Is this expected behavior? Could it be that uint8_t is an alias for some char-based type? See compiler/system notes in the code example.

// compile and run with:
// g++ test-uint8.cpp -std=c++11 && ./a.out
//                    -std=c++0x (for older gcc versions)
/**
 * prints out the following with compiler:
 *     gcc (GCC) 4.7.2 20120921 (Red Hat 4.7.2-2)
 * on the system:
 *     Linux 3.7.9-101.fc17.x86_64
 * Note that the first print statement uses an unset uint8_t
 * and therefore the behaviour is undefined. (Included here for
 * completeness)

> g++ test-uint8.cpp -std=c++11 && ./a.out
>>>�<<<    >>>194<<<
>>><<<    >>>0<<<
>>><<<    >>>0<<<
>>><<<    >>>0<<<
>>><<<    >>>1<<<
>>><<<    >>>2<<<

 *
 **/

#include <cstdint>
#include <iostream>

void print(const uint8_t& n)
{
    std::cout << ">>>" << n                 << "<<<    "
              << ">>>" << (unsigned int)(n) << "<<<\n";
}

int main()
{
    uint8_t a;
    uint8_t b(0);
    uint8_t c = 0;
    uint8_t d{0};
    uint8_t e = 1;
    uint8_t f = 2;
    for (auto i : {a,b,c,d,e,f})
    {
        print(i);
    }
}

推荐答案

code> uint8_t 是 unsigned char 的别名，iostreams对于打印字符而不是格式化数字的字符有特殊重载。

uint8_t is an alias for unsigned char, and the iostreams have special overloads for chars that print out the characters rather than formatting numbers.

转换为整数会禁止此操作。

The conversion to integer inhibits this.

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