How can I create a LLDB alias that evaluates its argument using `expression`

Question

I'm trying to create a LLDB alias that evaluates an expression using the argument (%1) to the alias. I've tried many, many different syntax combinations, but it seems like anything that uses %1 in an expression fails to parse.

(lldb) version
LLDB-112.1

This works as expected:

(lldb) expr (char*) strdup(argv[1])
(char *) $23 = 0x000000010061c090 "--calc"

When I create an alias containing %1 the example fails.

(lldb) command alias dup expr (char*) strdup(%1)
(lldb) dup argv[1]
error: expected expression
error: 1 errors parsing expression

How can I create a LLDB alias that evaluates its argument using expression?

Caslon

Answer 1

To make arguments in expression work, use command regex instead of alias, like this:

command regex dup 's/(.+)/expr (char*) strdup(%1)/'

For more details, see this answer.

Tavarez

Answer 2

I have played around with the %1 aliases quite a bit, no luck yet. This is the closest I can come, and it's not wonderful:

[.lldbinit]

command alias foo expr fooFunction(

Then when running the debugger, the following command will in fact call the function fooFunction with input bar:

(lldb) foo bar)

That unmatched close paren is not a typo; it needs to be there to match the open paren from the alias.

Oraleeoralia

Answer 3

try command alias dup expr -- (char*) strdup( $1 ); then you'll see

(lldb) dup "2387987"
(const char [8]) $5 = "2387987" {
  (const char) [0] = '2'
  (const char) [1] = '3'
  (const char) [2] = '8'
  (const char) [3] = '7'
  (const char) [4] = '9'
  (const char) [5] = '8'
  (const char) [6] = '7'
  (const char) [7] = '\0'

for lldb build 153

Chemisette

It looks like the $1 refers to something different than %1. Are you sure this is correct? I want the first argument to the macro, rather than the first evaluated expression output. – Caslon