I want to sample an external, possibly very short interrupt signal in my FPGA softcore. I did some research in some digital logic design books, and found this solution, where A is the input signal. The problem here is that it is way too short to be sampled by the synchronous flipflop shown, so a latch is used:
I am now wondering how this can be implemented in Verilog for use in a FPGA, I do not know how to tell the synthesis tool to place a latch there. How can this be done? Also, the design doesn't show how to reset Q - in the plot, it stays high for the rest of the example. How do I reset the whole design to be ready for the next interrupt signal?
Additionally: Do I need to put this signal through a few chained flip-flops to combat metastability, or is this design safe as-is?

