uint8_t 不能用 cout 打印 [英] uint8_t can't be printed with cout

问题描述

我有一个关于在 C++ 中处理整数的奇怪问题.

I have a weird problem about working with integers in C++.

我编写了一个简单的程序，为变量设置一个值，然后打印它，但它没有按预期工作.

I wrote a simple program that sets a value to a variable and then prints it, but it is not working as expected.

我的程序只有两行代码:

My program has only two lines of code:

uint8_t aa = 5;

cout << "value is " << aa << endl;

这个程序的输出是value is

即，它为 aa 打印空白.

I.e., it prints blank for aa.

当我将 uint8_t 更改为 uint16_t 时，上面的代码就像一个魅力.

When I change uint8_t to uint16_t the above code works like a charm.

我使用 Ubuntu 12.04 (Precise Pangolin)，64 位，我的编译器版本是:

I use Ubuntu 12.04 (Precise Pangolin), 64-bit, and my compiler version is:

gcc version 4.6.3 (Ubuntu/Linaro 4.6.3-1ubuntu5)

推荐答案

它并没有真正打印空白，但很可能是值为 5 的 ASCII 字符，它是不可打印的(或不可见的).有许多不可见的ASCII字符代码，其中大部分低于值32，实际上是空白.

It doesn't really print a blank, but most probably the ASCII character with value 5, which is non-printable (or invisible). There's a number of invisible ASCII character codes, most of them below value 32, which is the blank actually.

您必须将 aa 转换为 unsigned int 以输出数值，因为 ostream&operator<<(ostream&, unsigned char) 尝试输出可见字符值.

You have to convert aa to unsigned int to output the numeric value, since ostream& operator<<(ostream&, unsigned char) tries to output the visible character value.

uint8_t aa=5;

cout << "value is " << unsigned(aa) << endl;

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