[SOLVED] why is find's syntax so strange?

[mobile.Parmenides]

exec, a option of find command, takes a odd format as follows
find ... -exec command {} \;
Is there any explaination about why it takes this kind of format, or
some obscure implication?

[Cabhan]

What part of it do you find weird?

Basically, the way that -exec works is that it executes the command you specify for each file. So

find . -iname '*.html' -exec echo '{}' \;

might call:

echo index.html
echo contact.html

On the other hand, I might make a call like this:

find . -iname '*html' -exec my_program -r 3 '{}' \;

which would call:

my_program -r 3 index.html
my_program -r 3 contact.html

I assume that the strange bits are the "{}" and the "\;".

The "{}" is a special token that tells find where to put the name of the file that it found. It chose this symbol because it is very unlikely that a program would use this token, so find can safely assume that you want for it to use the {}.

The "\;" is because find uses a semicolon to terminate the command that you want to run. Therefore, note the differences between the following:

find . -iname '*.html' -exec echo '{}' -type f \;
find . -iname '*.html' -exec echo '{}' \; -type f

The first would print the "-type f" (assuming it to be an argument to echo), while the second treats "-type f" as an argument to find.

The reason for "\;" and not ";" is that Bash treats ";" as a special character. If you do not include the slash, Bash thinks that the ";" is meant for it (";" means to end the current command), and therefore find will not receive it, and will complain that you didn't use a semicolon:

alex@danu:~$ find . -iname '*.html' -exec echo '{}' ;
find: missing argument to `-exec'

Does this make sense?

[Rubberman]

What Cabhan said. Also, it is how the tool was originally written, in the deep dark early Unix era. It's been around for over 30 years now...

[mobile.Parmenides]

Your explaination is exactly what I need, thank.