Swift LLDB is not changing values / has no effect

Question

having a code like this:

var foo: Int = 0
print (foo)

With a breakpoint at the print line... then doing:

(lldb) exp foo = 7
(lldb) p foo
(Int) $R2 = 7
0

print(x) still returns 0 instead of 7. How is that possible?

Answer 1

More details would help, but in general, I have seen this happen in three cases:

1) When you have a let, the compiler is free to replace the let with its value. This is not your case, but for sake of argument, given:

let x = 0
print(x)

The compiler is free to rewrite this as

print(0)

leaving the let binding in only for the sake of the debugger. Now, a let is a let, so writing to it is not exactly a well-defined operation. With that said, C compilers seem to not take that same kind of approach with const values, so that failure mode might surprise somebody

2) If you call a function, the compiler is (essentially) going to emit code to push the arguments on the stack, and then jump to the code for the function. In theory, one would expect the line table in the debug info to guide the debugger to stop before the arguments are pushed. Then,

>> push x
call print

A write to x would actually "mean" something in the context of your print call. But, if the line tables actually make us stop at the call, even if you write to "x", it has already been pushed, and the old value will be used

3) The compiler might actually be using a copy of "foo" that is in a register, but for the sake of debug info, pointing us to a stack copy of "foo" which is actually never used, and only updated when writes to "foo" happen. If that is the case, LLDB will gladly write to the stack, but that won't affect the program's behavior as it is not actually reading from the stack

Fun things to try to figure this out include:

the LLDB disassemble command

(lldb) reg read pc
(lldb) dis -s <value of program counter>

or

(lldb) fr var -L foo

and then read at the memory address (or the register) that LLDB gives out as the location of "foo"

or

in your app, print(x) twice, and see if the second print shows a different value