Using floating point addition in verilog

Question

How can I represent floating point numbers in Verilog?

I am trying to use following code to do floating point addition but I am seeing integer as a output:

real r1,r2,r3;

initial begin
    r1 = 1.1;
    r2 = 1.2;
    r3 = r1+ r2;
    $display("Print the output %b", r3);
end

Qiu · Accepted Answer · 2014-12-26 18:12:04Z

1

%b indicates that a result should be print as binary. If you want to display a float number, you should use %e indicator.

$display("Print the output %e", r3);

Result before the change:

Print the output 00000000000000000000000000000010

and after:

Print the output 2.300000e+00

answered Dec 26, 2014 at 18:12

Qiu

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2 Comments

Thanks Qiu, Is there any way to print the output in 32bit binary form?

@naveenkumar: here you can find available format indicators.