doubt regarding blocking and non blocking statements in verilog?

[walkman]

could any one explain slid above slid?
please...

 

[kjm]

Note that all the statements will be executed whenever the value of "in" changes.

1). In o1=in; as soon as the value of in chnges, the same value is assigned to o1. So o1 is same as in.

2). In o2<=in; as soon as the value of "in" changes, its value is kept in a temporary variable and assigned to o2 at the end of the time.

3). #5 o3=in; In this case when value of in changes it enters into always block. Then waits for 5 time units and executes the command o3=in. So it follows the input till 30 time unit. After that when in=1, always block again executes,and waits for 5 time unit, but before 5 time unit (TU) gets over, the value of in changes to 0. So again it enters in always block and tries to wait for 5 TU, but before that again the value of in changes. Hence abrut changes between 30 to 40 TU are not traced to output.

4). #5 o4<=in; This case for this example is much similar to above point no. 3. However there will be difference in output if in changes its value once, and after 5 TU, again changes its value.

5). o5=#5 in; in this case, always block gets executed, and the statement also gets executed. But however, it waits for 5 TU, then takes the input and immediately assigns it to o5. So all the abrupt changes in time less than 5 TU are ignored.

6). o6 <= #5 in; In this case the value of in is taken immediately, and assignes to o6 after 5 TU. So output is the delayed version of input, where delay is 5 TU.

Comment for any clarifications.