I have a 256-bit value in Verilog:
reg [255:0] val;
I want to define a system task $foo that calls out to external C using the VPI, so I can call $foo like this:
$foo(val);
Now, in the C definition for the function 'foo', I cannot simply read the argument as an integer (PLI_INT32), because I have too many bits to fit in one of those. But, I can read the argument as a string, which is the same thing as an array of bytes. Here is what I wrote:
static int foo(char *userdata) {
vpiHandle systfref, args_iter, argh;
struct t_vpi_value argval;
PLI_BYTE8 *value;
systfref = vpi_handle(vpiSysTfCall, NULL);
args_iter = vpi_iterate(vpiArgument, systfref);
argval.format = vpiStringVal;
argh = vpi_scan(args_iter);
vpi_get_value(argh, &argval);
value = argval.value.str;
int i;
for (i = 0; i < 32; i++) {
vpi_printf("%.2x ", value[i]);
}
vpi_printf("\n");
vpi_free_object(args_iter);
return 0;
}
As you can see, this code reads the argument as a string and then prints out each character (aka byte) in the string. This works almost perfectly. However, the byte 00
always gets read as 20
. For example, if I assign the Verilog reg as follows:
val = 256'h000102030405060708090a0b0c0d0e0f101112131415161718191a1b1c1d1e1f;
And call it using $foo(val)
, then the C function prints this at simulation time:
VPI: 20 01 02 03 04 05 06 07 08 09 0a 0b 0c 0d 0e 0f 10 11 12 13 14 15 16 17 18 19 1a 1b 1c 1d 1e 1f
I have tested this with many different values and have found that the byte 00
always gets mapped to 20
, no matter where or how many times it appears in val
.
Also, note that if I read the value in as a vpiHexStrVal
, and print the string, it looks fine.
So, two questions:
- Is there a better way to read in my 256-bit value from the Verilog?
- What's going on with the
20
? Is this a bug? Am I missing something?
Note: I am using Aldec for simulation.