在cout文本中寻找 [英] seeking in cout text
问题描述
有没有办法在cout''d $ / b
之后跟踪x个字符，并覆盖部分字符？
is there any way to track back x number of characters after its been cout ''d
, and overwrite part of it ?
推荐答案
" Philip Parker" <博士** @ parker246.freeserve.co.uk>在消息中写道
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有什么方法可以跟踪x个字符后的字符数
''d，并覆盖部分字符？
is there any way to track back x number of characters after its been cout ''d , and overwrite part of it ?
否，至少不是标准C ++中的字符。您应该忘记cout和
而不是使用您的平台提供的任何控制台功能。这绝对是可能的，但不是标准的C ++。
john
No, at least not in standard C++.. You should probably forget about cout and
instead use whatever console functions your platform provides. This is
definitely possible, just not in standard C++.
john
>> ;有没有办法在cout
>> is there any way to track back x number of characters after its been cout
''d
之后跟踪x个字符，并覆盖部分字符？
, and overwrite part of it ?
<不，至少在标准C ++中没有。你应该忘记cout和
而是使用平台提供的任何控制台功能。这绝对是可能的，而不是标准的C ++。
No, at least not in standard C++.. You should probably forget about cout and
instead use whatever console functions your platform provides. This is
definitely possible, just not in standard C++.
也许使用一个继承自cout的新类？
int :: cout（char * s）{
real_cout<< s;
返回strlen（s）;
}
当然，像cout<< " BLA" << " FOO"那时候不行。
Jan Engelhardt
-
Maybe using a new class which inherits from cout?
int ::cout(char *s) {
real_cout << s;
return strlen(s);
}
Of course, something like cout << "bla" << "foo" won''t work then.
Jan Engelhardt
--
" Jan Engelhardt" < JE ***** @ linux01.gwdg.de>在消息中写道
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有没有办法在
$ b $之后跟踪x个字符数b cout''d
cout''d
，并覆盖它的一部分？
不，至少不是标准的C ++ ..你应该忘记cout
并且不使用任何控制台功能你的平台提供。这肯定是可能的，只是不是标准的C ++。
, and overwrite part of it ?
No, at least not in standard C++.. You should probably forget about cout andinstead use whatever console functions your platform provides. This is
definitely possible, just not in standard C++.
也许使用一个继承自cout的新类？
Maybe using a new class which inherits from cout?
cout是一个全局变量而不是一个类，所以你不能继承它。
int :: cout（char * s）{
real_cout<< s;
返回strlen（s）;
}
cout is a global variable not a class, so you can''t inherit from it.
int ::cout(char *s) {
real_cout << s;
return strlen(s);
}
这段代码对我没有意义，但无论如何你不要在cout中有'退后'的能力然后我不能看到你可以以某种方式获得
这种能力而不使用一些非标准的C ++ 。
john
That code makes no sense to me, but in any case if you don''t have the
ability to ''step back'' in cout then I can''t see that you could somehow get
that ability without using some non-standard C++.
john
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