Bourne/Bash shell special parameters: In POSIX shells, what does the special variable $* expand to during script execution?

Introduction / Context:
Shell scripts receive arguments as positional parameters. Special parameters like $, $@, $#, $?, , and $0 provide structured access to arguments, status codes, and process information. Understanding their semantics is critical for writing reliable scripts and interpreting command-line input.

Given Data / Assumptions:

• We are using a Bourne-compatible shell (sh, bash, dash, ksh, zsh in sh mode).
• Double-quoting behavior matters for $* versus $@.
• We need the definition of $* specifically.

Concept / Approach:

$* expands to all positional parameters. When double-quoted, "$*" expands to a single word containing all parameters joined by the first character of IFS (often a space). By contrast, "$@" expands to separate words, one per parameter, preserving argument boundaries. Other parameters serve different roles like exit status ($?), process ID (), and script name ($0).

Step-by-Step Solution:

Verification / Alternative check:

In a script invoked as ./s a "b c" d, echo "$" outputs a b c d (one word with embedded spaces as separated by IFS), while echo "$@" outputs a on one field, b c as one field, and d as another, preserving argument boundaries.

Why Other Options Are Wrong:

• Exit status: $? not $.
• Process ID: $$ not $.
• Command name: $0 not $.
• None of the above: Incorrect because the provided definition matches $.

Common Pitfalls:

Using "$" when you intend to forward arguments verbatim; prefer "$@" in that case to preserve argument separation. Be mindful of IFS changes, which alter how "$" joins arguments.

Final Answer:

list of all positional parameters as a single word