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

问题描述

我有一个奇怪的问题，使用整数在c ++。

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

我写了一个简单的程序，设置一个值到一个变量，然后打印它，不正常工作。

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;

此程序的输出为 value is

ie当我更改 uint8_t aa 输出空格。

i.e. it prints blank for aa.

> uint16_t 上面的代码就像一个charm。

when i change uint8_t to uint16_t the above code works like a charm.

我使用Ubuntu 12.04（64位）版本是： gcc版本4.6.3（Ubuntu / Linaro 4.6.3-1ubuntu5）

I use Ubuntu 12.04 (64 bit) and my compiler version is :gcc version 4.6.3 (Ubuntu/Linaro 4.6.3-1ubuntu5)

推荐答案

它不会真正打印一个空白，但最可能的ASCII字符值为5，这是不可见的。有许多不可见的ASCII字符代码，其中大多数都低于值32，这是

It doesn't really print a blank, but most probably the ASCII character with value 5, which is 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 输出数值，因为 ostream&

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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