Forum Discussion
2 Replies
- Altera_Forum
Honored Contributor
In that particular example it *should* be fine, because it is effectively equivalent to:
A is still used as the clock, and D is used as a clock enable.wire rst_n = !(B & C); always @ (posedge A or negedge rst_n) begin if(~rst_n) begin E <= 0; end else if (D) begin E <= F; end end - Altera_Forum
Honored Contributor
--- Quote Start --- In that particular example it *should* be fine, because it is effectively equivalent to:
A is still used as the clock, and D is used as a clock enable. --- Quote End --- Thank you. After post this, I try use Quartus "Technology Map Viewer", and it give me the answer. Thank you anyway, you're very kind.wire rst_n = !(B & C); always @ (posedge A or negedge rst_n) begin if(~rst_n) begin E <= 0; end else if (D) begin E <= F; end end