Ok, I am a bit embarrassed to ask such a simple thing but still.
好吧，我仍然有点尴尬地问这么简单的事情。
I have command line utility application and need to show progress to the user.
我有命令行实用程序应用程序，需要向用户显示进度。
I could write progress into cout, like this:
我可以写进cout的进度，像这样：
std::cout << "10%\n";
...
std::cout << "20%\n";
...
std::cout << "30%\n";
... but as a result user will see:
...但结果用户会看到：
some line printed before
10%
20%
30%
...
... but what i really need is that percentage got updated, like this at the beginning:
...但我真正需要的是百分比得到更新，如下所示：
some line printed before
10%
...
... and after update:
......并在更新后：
some line printed before
20%
...
... and after second update:
...并在第二次更新后：
some line printed before
30%
...
How should I achieve that?
我该怎么做？
2 个解决方案
#1
24
Instead of using '\n', use '\r':
不使用'\ n'，而是使用'\ r'：
std::cout << "\r10%" << std::flush;
Print newline ('\n') when done.
完成后打印换行符（'\ n'）。
It's important to use std::flush so the stream contents really is output.
使用std :: flush非常重要，因此输出了流内容。
#2
6
Use a carriage return.
使用回车。
std::cout << "\r10%";
std::cout << "\r20%";
...
Goes to the beginning of the line.
走到行的开头。
#1
24
Instead of using '\n', use '\r':
不使用'\ n'，而是使用'\ r'：
std::cout << "\r10%" << std::flush;
Print newline ('\n') when done.
完成后打印换行符（'\ n'）。
It's important to use std::flush so the stream contents really is output.
使用std :: flush非常重要，因此输出了流内容。
#2
6
Use a carriage return.
使用回车。
std::cout << "\r10%";
std::cout << "\r20%";
...
Goes to the beginning of the line.
走到行的开头。