Operator &~

SystemVerilog
1

module nand_1();
logic a,b;
logic y;
initial
begin
a=0;
b=0;
y= a&~b;
$display(“y=%b”,y);

end

endmodule

after run

y=0 ?

this is like

module nand_1();
logic a,b;
logic y,y1;
initial
begin
a=0;
b=0;
y= a&~b;
y1=a&b;
//y= ~(a&~b);
$display(“y=%b,y1=%b”,y,y1);

end
endmodule
after run
y=0,y1=0
both are same but using different operators &, &~

as per NAND logic , if both inputs are 0, the output will be 1

please help me

2

In reply to suresh M:

You have the characters reversed. ~& is a nand.

3

In reply to dave_59:

module nand_1();
logic a,b;
logic y;
initial
begin
a=0;
b=0;
y= ~&b;//y=b ~& b;
$display(“y=%b”,y);

end
endmodule

result:
y=1

This is like y=b ~& b;
for y= a~& b;

after run
Invalid use of unary operator ‘~&’
“testbench.sv”, 10: token is ‘;’
y= a~&b;
^

4

In reply to suresh M:

Dear suresh,

~ is a bitwise operator, it requires one operand.
~& b is the same as ~b no matter the value of b (N.B: if b is one bit wide)

The tool translates your expression y = b ~& b; as y = b (~&b); which has no sense.

5

In reply to suresh M:

If you are trying to write the RTL equivalent for

nand ( y, a, b);

that would be

y = !(a && b);
// or
y = !a || !b;
6

In reply to dave_59:

then what is the use &~ and &~

if I use
a=0;b=0;
y=a&~b;
result is y=0

but when
a=1;
b=0;
y = a&~b;

result is y=1

otherway if I use
y=a~&b;

after run
Invalid use of unary operator ‘~&’
“testbench.sv”, 10: token is ‘;’
y= a~&b;

7

In reply to vico:

I understood that ~& is unary operator like either ~b or ~&b , both are giving complement of b.

but my doubt is when we use &~ operator
example
a=0;
b=0;
y=a&~b;

what is y and how?

8

In reply to vico:

I understood that ~& is unary operator like either ~b or ~&b , both are giving complement of b.

but my doubt is when we use &~ operator
example
a=0;
b=0;
y=a&~b;

what is y and how?

9

In reply to suresh M:

There is no single &~ operator. As Vico mentioned above

y = a&~b;

is interpreted as two distinct operators

y = a & ~b ;

If you wrote

int v = 5*-3;

v would be -15. There is no *- operator.

10

In reply to dave_59:

thank you so much