VBA expression evaluation order - vba

Question

I wanted to find out in which order the VBA evaluates the terms inside the expression.
I have read the document about operator precedence and thought it will simply evaluate from left to right if the terms are all operated with "+".
However I get these strange results.
Option Explicit
Dim x
Sub test()
Debug.Print CreateObject("Scripting.FileSystemObject") _
.GetFileVersion(Application.Path & "\Excel.exe") ' 16.0.12624.20086
x = 0: Debug.Print inc_x + x '(a) 2
x = 0: debug_print inc_x + x '(b) 2
x = 0: Debug.Print x + inc_x '(c) 2 <=== ?
x = 0: debug_print x + inc_x '(d) 2 <=== ?
x = 0: Debug.Print inc_x + 0 + x '(e) 2
x = 0: debug_print inc_x + 0 + x '(f) 2
x = 0: Debug.Print 0 + x + inc_x '(g) 2 <=== ?
x = 0: debug_print 0 + x + inc_x '(h) 1 <=== ???
End Sub
Function inc_x()
x = x + 1
inc_x = x
End Function
Function debug_print(msg)
Debug.Print msg
End Function
I thought (c), (d), and (g) should print 1 but they print 2.
And it gets even weirder when you look at (h) which is the only one printing 1.
Can someone explain what is happening ?
added more patterns
I added (i),(j),(k),(l),(m), and (n) which also shows interesting patterns.
#SJR also taught me that specifying the type of x as Long will change the result of (d).
Option Explicit
Dim x 'As Long 'specifying the type will change the result of (d)
Sub test()
Debug.Print CreateObject("Scripting.FileSystemObject") _
.GetFileVersion(Application.Path & "\Excel.exe") ' 16.0.12624.20086
x = 0: Debug.Print inc_x + x '(a) 2
x = 0: debug_print inc_x + x '(b) 2
x = 0: Debug.Print x + inc_x '(c) 2 <=== ?
x = 0: debug_print x + inc_x '(d) 2 <=== ? (This becomes 1 when x is Long)
x = 0: Debug.Print inc_x + 0 + x '(e) 2
x = 0: debug_print inc_x + 0 + x '(f) 2
x = 0: Debug.Print 0 + x + inc_x '(g) 2 <=== ?
x = 0: debug_print 0 + x + inc_x '(h) 1 <=== ???
x = 0: Debug.Print 100 * inc_x + 10 * x + 1 * inc_x '(i) 122
x = 0: debug_print 100 * inc_x + 10 * x + 1 * inc_x '(j) 112
x = 0: Debug.Print 1000 * x + 100 * inc_x + 10 * x + 1 * inc_x '(k) 2122
x = 0: debug_print 1000 * x + 100 * inc_x + 10 * x + 1 * inc_x '(l) 112
x = 0: Debug.Print 10000 * inc_x + 1000 * x + 100 * inc_x + 10 * x + 1 * inc_x '(m) 13233
x = 0: debug_print 10000 * inc_x + 1000 * x + 100 * inc_x + 10 * x + 1 * inc_x '(n) 11223
End Sub
Function inc_x()
x = x + 1
inc_x = x
End Function
Function debug_print(msg)
Debug.Print msg
End Function