需要帮助理解\\ n \\ b和\\ R将如何​​渲染输出的printf [英] Need help understanding how \n, \b, and \r will render printf output

问题描述

我写在C下面的程序，当我运行它，我被看的输出感到惊讶。

I wrote the following program in the C and when I run it, I was surprised by looking at the output.

下面是节目

int main()
{    
       printf("\nab");
       printf("\bsi");    
       printf("\rha");    
}

输出是： - 海
而我期待absiha，因为\\ n是新行，\\ b是退格（非擦除）和\\ r表示回车。所以我期待的柯森是在我字，因为\\ R已经被应用，但是当我运行它，看到输出我完全感到惊讶和困惑。任何人都可以请解释一下我的输出？

The output is :- hai whereas I was expecting "absiha" since \n is for new line, \b is for backspace(non erase) and \r is for carriage return. So I was expecting that curson would be at "i" character because \r has been applied but when I run it and saw the output I was totally surprised and confused. Can anyone please explain me the output?

推荐答案

让我们把它一步步时间：

Let's take it one step at a time:

＆LT;新行＆GT; AB＆LT;退格＆GT;硅＆lt;回车＆GT;哈

首先，办理退格。注意，即使它是非擦除，下一个字符被输出将覆盖什么退格以上：

First, handle the backspace. Note that even though it is "non-erase", the next character to be output would overwrite what was backspaced over:

＆LT;新行＆GT; ASI＆LT;回车＆GT;哈

现在，回车装置返回该行的开头。因此，哈覆盖为ASI：

Now, a carriage return means to go back to the beginning of the line. So the "ha" overwrites the "as" in "asi:

＆LT;新行＆GT;海

现在，游标当前坐在 I ，所以接下来的字符输出会覆盖 I

Now, the cursor is currently sitting on the i, so the next character to be output would overwrite i.

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