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Signals of type wire or a similar wire like data type requires the continuous assignment of a value. For example, consider an electrical wire used to connect pieces on a breadboard. As long as the +5V battery is applied to one end of the wire, the component connected to the other end of the wire will get the required voltage.


In Verilog, this concept is realized by the an assign statement where any wire or other similar wire like data-types can be driven continuously with a value. The value can either be a constant or an expression comprising of a group of signals.

Assign Syntax

The assignment syntax starts with the keyword assign followed by the signal name which can be either a single signal or a concatenation of different signal nets. The drive strength and delay are optional and are mostly used for dataflow modeling than synthesizing into real hardware. The expression or signal on the right hand side is evaluated and assigned to the net or expression of nets on the left hand side.

assign <net_expression> = [drive_strength] [delay] <expression of different signals or constant value>


Delay values are useful for specifying delays for gates and are used to model timing behavior in real hardware because the value dictates when the net should be assigned with the evaluated value.


There are some rules that need to be followed when using an assign statement:

  • LHS should always be a scalar or vector net or a concatenation of scalar or vector nets and never a scalar or vector register.
  • RHS can contain scalar or vector registers and function calls.
  • Whenever any operand on the RHS changes in value, LHS will be updated with the new value.
  • assign statements are also called continuous assignments and are always active

Example #1

In the following example, a net called out is driven continuously by an expression of signals. i1 and i2 with the logical AND & form the expression.


If the wires are instead converted into ports and synthesized, we will get an RTL schematic like the one shown below after synthesis.

Continuous assignment statement can be used to represent combinational gates in Verilog.

Example #2

The module shown below takes two inputs and uses an assign statement to drive the output z using part-select and multiple bit concatenations. Treat each case as the only code in the module, else many assign statements on the same signal will definitely make the output become X.

module xyz (input [3:0] 	x,		// x is a 4-bit vector net
						input 				y, 		// y is a scalar net (1-bit)
						output [4:0] 	z ); 	// z is a 5-bit vector net

wire [1:0] 	a;
wire 				b;
// Assume one of the following assignments are chosen in real design
// If x='hC and y='h1 let us see the value of z  

// Case #1: 4-bits of x and 1 bit of y is concatenated to get a 5-bit net
// and is assigned to the 5-bit nets of z. So value of z='b11001 or z='h19
assign z = {x, y};

// Case #2: 4-bits of x and 1 bit of y is concatenated to get a 5-bit net
// and is assigned to selected 3-bits of net z. Remaining 2 bits of z remains
// undriven and will be high-imp. So value of z='bZ001Z
assign z[3:1] = {x, y};

// Case #3: The same statement is used but now bit4 of z is driven with a constant
// value of 1. Now z = 'b1001Z because only bit0 remains undriven
assign z[3:1] = {x, y};
assign z[4] = 1;

// Case #4: Assume bit3 is driven instead, but now there are two drivers for bit3,
// and both are driving the same value of 0. So there should be no contention and 
// value of z = 'bZ001Z
assign z[3:1] = {x, y};
assign z[3] = 0;

// Case #5: Assume bit3 is instead driven with value 1, so now there are two drivers
// with different values, where the first line is driven with the value of X which
// at the time is 0 and the second assignment where it is driven with value 1, so
// now it becomes unknown which will win. So z='bZX01Z
assign z[3:1] = {x, y};
assign z[3] = 1;

// Case #6: Partial selection of operands on RHS is also possible and say only 2-bits
// are chosen from x, then z = 'b00001 because z[4:3] will be driven with 0
assign z = {x[1:0], y};

// Case #7: Say we explicitly assign only 3-bits of z and leave remaining unconnected
// then z = 'bZZ001
assign z[2:0] = {x[1:0], y};

// Case #8: Same variable can be used multiple times as well and z = 'b00111
// 3{y} is the same as {y, y, y}
assign z = {3{y}};

// Case #9: LHS can also be concatenated: a is 2-bit vector and b is scalar
// RHS is evaluated to 11001 and LHS is 3-bit wide so first 3 bits from LSB of RHS
// will be assigned to LHS. So a = 'b00 and b ='b1
assign {a, b} = {x, y};

// Case #10: If we reverse order on LHS keeping RHS same, we get a = 'b01 and b='b0
assign {a, b} = {x, y};



Assign reg variables

It is illegal to drive or assign reg type variables with an assign statement. This is because a reg variable is capable of storing data and does not require to be driven continuously. reg signals can only be driven in procedural blocks like initial and always.

Implicit Continuous Assignment

When an assign statement is used to assign the given net with some value, it is called explicit assignment. Verilog also allows an assignment to be done when the net is declared and is called implicit assignment.

wire [1:0] a;
assign a = x & y; 			// Explicit assignment

wire [1:0] a = x & y; 	// Implicit assignment


Combinational Logic Design

Consider the following digital circuit made from combinational gates and the corresponding Verilog code.


Combinational logic requires the inputs to be continuously driven to maintain the output unlike sequential elements like flip flops where the value is captured and stored at the edge of a clock. So an assign statement fits the purpose the well because the output o is updated whenever any of the inputs on the right hand side change.

// This module takes four inputs and performs a boolean
// operation and assigns output to o. The combinational
// logic is realized using assign statement.

module combo (	input 	a, b, c, d,
								output  o);
  assign o = ~((a & b) | c ^ d);


Hardware Schematic

After design elaboration and synthesis, we do get to see a combinational circuit that would behave the same way as modeled by the assign statement.

combinational gate schematic

See that the signal o becomes 1 whenever the combinational expression on the RHS becomes true. Similarly o becomes 0 when RHS is false. Output o is X from 0ns to 10ns because inputs are X during the same time.


Click here for a slideshow with simulation example !

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