To determine the contents of the itab
internal table after executing the given code, let's go step by step:
Move: 'A' to itab-field1, 'B' to itab-field2.
- This moves the values 'A' and 'B' to itab-field1
and itab-field2
respectively. So, itab
now contains one entry with field1
as 'A' and field2
as 'B'.
Append itab.
- This appends the current contents of itab
to itab
. So, itab
now contains two entries with field1
as 'A' and field2
as 'B'.
Append itab.
- This appends the current contents of itab
to itab
again. So, itab
now contains three entries with field1
as 'A' and field2
as 'B'.
Move: 'B' to itab-field1.
- This moves the value 'B' to itab-field1
. So, the first entry in itab
now has field1
as 'B' and field2
as 'B'.
Append itab.
- This appends the current contents of itab
to itab
. So, itab
now contains four entries with field1
as 'B' and field2
as 'B'.
Clear itab.
- This clears the contents of itab
.
Move: 'A' to itab-field2.
- This moves the value 'A' to itab-field2
. So, the first entry in itab
now has field1
as 'B' and field2
as 'A'.
Append itab.
- This appends the current contents of itab
to itab
. So, itab
now contains one entry with field1
as 'B' and field2
as 'A'.
Therefore, the contents of itab
after executing the code are:
A B B A
So, the correct answer is A.