Question

我的“测试”文件有3种空行：

$ cat -nA test
     1  This is line 1 and it's followed by a blank line$
     2  $
     3  This is line 2 and it's followed by a 'blank' line with 3 tab characters$
     4  ^I^I^I$
     5  This is line 3 and it's followed by a 'blank' line with 3 whitespace characters$
     6     $
     7  This is line 4$

如果我用POSIX字符类“blank”或“space”开头的空行，我会得到以制表符或空格开头的空行：

$ grep -n '^[[:blank:]]' test
4:          
6:   
$ grep -n '^[[:space:]]' test
4:          
6:

如果我反转搜索，我会得到所有其他行：

$ grep -nv '^[[:blank:]]' test
1:This is line 1 and it's followed by a blank line
2:
3:This is line 2 and it's followed by a 'blank' line with 3 tab characters
5:This is line 3 and it's followed by a 'blank' line with 3 whitespace characters
7:This is line 4

$ grep -nv '^[[:space:]]' test
1:This is line 1 and it's followed by a blank line
2:
3:This is line 2 and it's followed by a 'blank' line with 3 tab characters
5:This is line 3 and it's followed by a 'blank' line with 3 whitespace characters
7:This is line 4

然而，如果我否定搜索，我只得到非空行：

$ grep -n '^[^[:blank:]]' test
1:This is line 1 and it's followed by a blank line
3:This is line 2 and it's followed by a 'blank' line with 3 tab characters
5:This is line 3 and it's followed by a 'blank' line with 3 whitespace characters
7:This is line 4

$ grep -n '^[^[:space:]]' test
1:This is line 1 and it's followed by a blank line
3:This is line 2 and it's followed by a 'blank' line with 3 tab characters
5:This is line 3 and it's followed by a 'blank' line with 3 whitespace characters
7:This is line 4

为什么否定否定与反转相同？

Answer 1

[^[:space:]]表示“除[:space:]类之外的任何字符”。因此，匹配^[^[:space:]]的行必须至少包含一个字符。

在这种情况下，为什么grep反演与否定不一样？

1 个答案: