linux shell: for arg; do

Question

Reading LTP shell code it uses strange for loop syntax:

for arg; do
    TCID="${TCID}_$arg"
done

1. How does it takes arguments? I'd expect it loops over $arg, separating with $IFS, but when trying $ arg="aa bb"; for arg; do echo $arg; done and it prints nothing.
2. Is it a bashism?

Answer 1

It's a special way of handling command-line options using for loop.

It's equivalent to:

for arg in "$@"; do
    TCID="${TCID}_$arg"
done

It's not specific to bash. It's defined in POSIX:

The for Loop

The for loop shall execute a sequence of commands for each member in a list of items. The for loop requires that the reserved words do and done be used to delimit the sequence of commands.

The format for the for loop is as follows:

for name [ in [word ... ]] do  
   compound-list  
done

First, the list of words following in shall be expanded to generate a list of items. Then, the variable name shall be set to each item, in turn, and the compound-list executed each time. If no items result from the expansion, the compound-list shall not be executed. Omitting:

in word ...

shall be equivalent to:

in "$@"