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Example:

Input:

0x12345678 0aef1234 0098adefa 123456789 

Expected Output:

0x00000000 00000000 0098adefa 123456789

Tried this:

sed -E "s/[0-9a-fA-F]8/00000000/g" 

But this replaces even if there are more than 8 continues hex digits.

You can use word boundary b to mark the beginning or the end of a word, and additionally one needs to take into account that the 0x is possible in front of the number:

sed -E 's/b(0x|)[0-9a-fA-F]8b/100000000/g'

Still, there are special cases which you need to think through what to do with them, like 00x12345678, in the above pattern it is not replaced.

In Perl, you can use the "look-around assertions", i.e. you can say "not preceded by a hexdigit" and "not followed by a hexdigit":

perl -pe 's/(?<![0-9a-fA-F])[0-9a-fA-F]8(?![0-9a-fA-F])/00000000/g'

The look-arounds don't count into the matched string, i.e. they aren't replaced and they can match more than once (once in the look-behind and once in the look-ahead).

You can also use the [:xdigit:] POSIX class instead of 0-9a-fA-F.

$ echo '0x12345678 0aef1234 0098adefa 123456789' | sed -E 's/(<|x)[[:xdigit:]]8>/100000000/g'
0x00000000 00000000 0098adefa 123456789

$ echo '12345678 0aef1234 0098adefa 123456789' | sed -E 's/(<|x)[[:xdigit:]]8>/100000000/g'
00000000 00000000 0098adefa 123456789

The regular expression

(<|x)[[:xdigit:]]8>

will match a hexadecimal number that is eight digits long. It should be a complete word, or be preceded by an x. The preceding character (the x, if there is one) is saved and inserted before the zeros in the replacement.