while IFS= read -r -d $'\0' file ... explanation
bash
Solution
In Bash, `varname=value command` runs command with the environment variable `varname` set to `value` (and all other environment variables inherited normally). So `IFS= read -r -d $'\0'` runs the command `read -r -d $'\0'` with the environment variable `IFS` set to the empty string (meaning no field separators).
Since `read` returns success (i.e., sets `$?` to `0`) whenever it successfully reads input and doesn't encounter end-of-file, the overall effect is to loop over a set of NUL-separated records (saved in the variable `REPLY`).
Doesn't the while statement need a 'test' or [ ] or [[ ]] expression that will set $? to 1 or 0?
`test` and `[ ... ]` and `[[ ... ]]` aren't actually expressions, but commands. In Bash, every command returns either success (setting `$?` to `0`) or failure (setting `$?` to a non-zero value, often `1`).
(By the way, as nosid notes in a comment above, `-d $'\0'` is equivalent to `-d ''`. Bash variables are internally represented as C-style/NUL-terminated strings, so you can't really include a NUL in a string; e.g., `echo $'a\0b'` just prints `a`.)
Problem
I do not understand this line of shell script. Doesn't the while statement need a 'test' or [ ] or [[ ]] expression that will set $? to 1 or 0? How does ``` while IFS= read -r -d $'\0'; do ...; done ``` do that? Any help understanding the syntax here is greatly appreciated.