CODEWITHSUNDEEP
objective-cswiftlldb
rustyMagnet
rustyMagnet

Reputation: 4095

Create Swift class instance variable with lldb, versus Objective-C

Using my debugger (lldb) I can easily create Instances classes when it is Objective-C code. 

(lldb) e id $my_hello = [hello_from_objc new]
(lldb) po $my_hello
<hello_from_objc: 0x1c4013020>
(lldb) po [$my_hello secret_objc_method]
0x000000000000002a
(lldb) po (int) [$my_hello secret_objc_method]
42

But I can't work out how to do the same with lldb's expression command when the code is pure Swift. I create an instance in Swift code easily enough..

let my_swift_framework = Hello_Class()
print("✔️ \(my_swift_framework.samplePublicVariable)")

Upvotes: 4

Views: 788

Answers (1)

Martin R
Martin R

Reputation: 539945

Here is an example: After executing the Swift code

class HelloClass {
    func hello() {
        print("hello")
    }
}

you can create an object in the debugger window:

(lldb) expression let $myHello = HelloClass()
(lldb) po $myHello
<hello_class: 0x101121180>

(lldb) po $myHello.hello()
hello

If you get an error

error: unknown type name 'let'
error: use of undeclared identifier 'HelloClass'

then set the expression language to Swift explicitly:

(lldb) expression -l swift -o -- let $myHello = HelloClass()

(lldb) expression -l swift -o -- $myHello.hello()

Or you can change the lldb language context back to Swift:

(lldb) settings set target.language swift

Upvotes: 7

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