sed 替换字符串

Question

(declare-const buabor Real)

which I want to replace as我想替换为

(declare-fun buabor () Real)

can it be done with sed?可以用 sed 完成吗？

I tried我试过了

sed 's/(declare-const\s([a-z])\s(Real))/(declare-fun\s\2\(\)\3/)g'

but was not able to get the result any help would be great但无法得到结果任何帮助都会很棒

Answer 1

This should do (assuming GNU sed as \s is used):这应该可以（假设GNU sed作为\s ）：

sed 's/(declare-const\s\([a-z]*\)/(declare-fun \1()/'

By default, BRE is used, so ( and ) will be matched literally in search section.默认情况下，使用 BRE，因此(和)将在搜索部分按字面意思匹配。 See also BRE-vs-ERE section in the manual另请参阅手册中的BRE-vs-ERE部分
\( and \) will then become capture group \(和\)然后将成为捕获组
[az]* will match zero or more lowercase alphabets [az]*将匹配零个或多个小写字母
rest of the line need not be matched and used in replacement section as it isn't modified该线路的rest未经过修改，无需匹配并用于替换部分

A few more observations:还有一些观察：

( and ) aren't special in replacement section (和)在替换部分并不特殊
\s is again not special in replacement section, will insert s \s在替换部分再次不是特殊的，将插入s