Verilog Procedural Interface - Example

Example

As an example, consider the following Verilog code fragment:

val = 41; $increment(val); $display("After $increment, val=%d", val);

Suppose the increment system task increments its first parameter by one. Using C and the VPI mechanism, the increment task can be implemented as follows:

// Implements the increment system task static int increment(char *userdata) { vpiHandle systfref, args_iter, argh; struct t_vpi_value argval; int value; // Obtain a handle to the argument list systfref = vpi_handle(vpiSysTfCall, NULL); args_iter = vpi_iterate(vpiArgument, systfref); // Grab the value of the first argument argh = vpi_scan(args_iter); argval.format = vpiIntVal; vpi_get_value(argh, &argval); value = argval.value.integer; vpi_printf("VPI routine received %d\n", value); // Increment the value and put it back as first argument argval.value.integer = value + 1; vpi_put_value(argh, &argval, NULL, vpiNoDelay); // Cleanup and return vpi_free_object(args_iter); return 0; }

Also, a function that registers this system task is necessary. This function is invoked prior to elaboration or resolution of references when it is placed in the externally visible vlog_startup_routines array.

// Registers the increment system task void register_increment { s_vpi_systf_data data = {vpiSysTask, 0, "$increment", increment, 0, 0, 0}; vpi_register_systf(&data); } // Contains a zero-terminated list of functions that have to be called at startup void (*vlog_startup_routines) = { register_increment, 0 };

The C code is compiled into a shared object that will be used by the Verilog simulator. A simulation of the earlier mentioned Verilog fragment will now result in the following output:

VPI routine received 41 After $increment, val=42