bash - The "non-man" explanation of -n in echo (shell)

Question

So i read the man echo info. And it says this for the n option.

-n    Do not print the trailing newline character.

I can't wrap my head around what it means.

So if i try it out this is what i get

~ echo -n "12"
12%

I get a % after the argument I passed in.

Without n, stdout will not have %

~ echo "12"
12

Can anyone give practical examples, or a simpler explanation.

Answer 1

命令的标准调用echo将打印出传递给的参数echo，后跟换行符。

echo使用标志调用-n只会打印出参数，没有换行符。

因为您%在 shell 中使用了提示符，所以调用echo -n "12", 将打印出12，紧随其后的提示符表示 shell 已准备好接受新输入。

如果没有：它看起来像这样-n：

% echo "12"
12
%

并与-n：

% echo -n "12"
12%