为什么“cout"?对“无符号字符"工作很奇怪? [英] Why "cout" works weird for "unsigned char"?

问题描述

我有以下代码:

cvtColor (image, image, CV_BGRA2RGB);
Vec3b bottomRGB;
bottomRGB=image.at<Vec3b>(821,1232);

当我显示bottomRGB[0]时，显示的值大于255，这是什么原因?

When I display bottomRGB[0], it displays a value greater than 255. What is the reason for this?

推荐答案

正如你所评论的，原因是你使用 cout 直接打印其内容.在这里，我将尝试向您解释为什么这行不通.

As you have commented, the reason is that you use cout to print its content directly. Here I will try to explain to you why this will not work.

cout << bottomRGB[0] << endl;

为什么 "cout" 对 "unsigned char" 工作很奇怪?

它不会工作，因为这里 bottomRGB[0] 是一个 unsigned char(值为 218)，cout 实际上会打印一些垃圾值(或什么都不打印)，因为它只是一个不可打印 ASCII 字符，无论如何都会打印出来.请注意，对应于 218 的 ASCII 字符是不可打印的.在此处查看 ASCII 表.

Why "cout" works weird for "unsigned char"?

It will not work because here bottomRGB[0] is a unsigned char (with value 218), cout actually will print some garbage value (or nothing) as it is just a non-printable ASCII character which is getting printed anyway. Note that ASCII character corresponding to 218 is non-printable. Check out here for the ASCII table.

附言您可以使用 bottomRGB[0] 是否可打印>isprint() 为:

P.S. You can check whether bottomRGB[0] is printable or not using isprint() as:

cout << isprint(bottomRGB[0]) << endl; // will print garbage value or nothing

它将打印0(或false)表示字符不可打印

It will print 0 (or false) indicating the character is non-printable

对于您的示例，要使其工作，您需要在 cout 之前先键入 cast:

For your example, to make it work, you need to type cast it first before cout:

cout << (int) bottomRGB[0] << endl; // correctly printed (218 for your example) 

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