Count the number of closed inputs in module 0.
In this exercise we will use a shift operation to apply a solution other than
that used in the previous example.
The counter for closed inputs is still MB80 and its initial value is 0 at the
beginning of the OB1. MB10 represents a mask of 8 bits only one of which will,
in turn, be set to 1. The initial value is 1, that is (00000001)2, where only
bit 0 is true.
The first three statements of the FB4 block load the accumulators with the
values of the mask and the byte corresponding to the input to module 0 and then
execute the AND one bit at a time. On the first step, with the value of the mask
hardly visible, the AND only gives a result other than 0 if E0.0 is closed. This
means that the derivation on zero included in the next statement will only be
carried out if the input is open.
If the input is closed, as in the previous example, the counter byte MB80.0 is
increased.
Whatever the case, the two paths come together in label INC where, after loading
the mask in ACCU1, it is shifted one position to the left. The result of this
displacement, or better yet the low part, is again transferred to MB10. After
the statement L KF+256, ACCU1 will have the value of 256 and ACCU2 will contain
the result of the displacement. If the mask bit, after 8 displacements, has
ended up in position 8 of the 16 bit register of the accumulator, the value of
this accumulator will be 28 = 256, the statement of the conditional jump is
ignored and processing of the block stops. If the bit is located in lower
positions then the content of the accumulator is less than 256 and the jump to
the TEST label is executed to repeat the process and examine the next input.
More examples: [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] [ 6 ] [ 7 ] [ 8 ] [ 9 ] [ 10 ] [ 11 ] [ 12 ] [ 13 ] [ 14 ] [ 15 ] [ 16 ] [ 17 ] [ 18 ] [ 19 ] [ 20 ] [ 21 ] [ 22 ] [ 23 ] [ 24 ] [ 25 ] [ 26 ] [ 27 ] [ 28 ] [ 29 ] [ List
]
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