Table transformation, table as list of dicts - kdb

Question

Please help me with t1 transformation through the following path:
t1:enlist `a`b!1 2;
t2:exec val from ([]val:t1);
-3!t1 // +`a`b!(,1;,2)
-3!t2 // ,`a`b!1 2
t1~t2[;] // 1b
I expect that the 2nd line (exec) returns the same object as t1, but it is not. For some reason only [;] gets t1 from t2.
So what (and why) happens on lines 2 and 5?
UPD
Why the enlist is so fruitful? It makes enlist for each element, and also flips the entire object
-3!enlist `a`b!1 2 // +`a`b!(,1;,2)
-3!enlist each `a`b!1 2 // `a`b!(,1;,2)
-3!flip enlist each `a`b!1 2 // +`a`b!(,1;,2)

Answer 1

I think what's happening here is that the special promotion of lists of (conforming) dictionaries to a table is essentially "undone" when it becomes the column of a table. It goes back to being lists of dictionaries as if they were non-conforming, though misleadingly the terminal still displays it as if it was a table. To see this, consider this example:
/promoted to table
q)-3!(`a`b!4 5;`a`b!1 2)
"+`a`b!(4 1;5 2)"
/still a table (embedded)
q)-3!enlist(`a`b!4 5;`a`b!1 2)
",+`a`b!(4 1;5 2)"
/still a table (embedded)
q)-3!(1#`val)!enlist(`a`b!4 5;`a`b!1 2)
"(,`val)!,+`a`b!(4 1;5 2)"
/no longer a table, now a list of dictionaries
q)-3!flip(1#`val)!enlist(`a`b!4 5;`a`b!1 2)
"+(,`val)!,(`a`b!4 5;`a`b!1 2)"
/extracting the column gives an "un-promoted" list of dictionaries
q)-3!exec val from flip(1#`val)!enlist(`a`b!4 5;`a`b!1 2)
"(`a`b!4 5;`a`b!1 2)"
/even though there are two of them and they both have type 99h, it is not a table
q)count exec val from flip(1#`val)!enlist(`a`b!4 5;`a`b!1 2)
2
q)type each exec val from flip(1#`val)!enlist(`a`b!4 5;`a`b!1 2)
99 99h
q)type exec val from flip(1#`val)!enlist(`a`b!4 5;`a`b!1 2)
0h
/so it is not comparable to something that has the promotion, even though the terminal makes them appear the same
q)((`a`b!4 5;`a`b!1 2))~exec val from flip(1#`val)!enlist(`a`b!4 5;`a`b!1 2)
0b
q)(`a`b!4 5;`a`b!1 2)
a b
---
4 5
1 2
q)exec val from flip(1#`val)!enlist(`a`b!4 5;`a`b!1 2)
a b
---
4 5
1 2
Your case is the same thing but for a single dictionary rather than two, but I think two dictionaries makes it easier to see.