assign verilog conditional

Verilog Conditional Statements

In Verilog, conditional statements are used to control the flow of execution based on certain conditions. There are several types of conditional statements in Verilog listed below.

Conditional Operator

The conditional operator allows you to assign a value to a variable based on a condition. If the condition is true, expression_1 is assigned to the variable. Otherwise, expression_2 is assigned.

Nested conditional operators

Conditional operators can be nested to any level but it can affect readability of code.

Here are some of the advantages of using conditional operators:

Concise syntax : The conditional operator allows for a compact and concise representation of conditional assignments. It reduces the amount of code needed compared to using if-else statements or case statements.
Readability : The conditional operator can enhance code readability, especially for simple conditional assignments. It clearly expresses the intent of assigning different values based on a condition in a single line.

And some disadvantages:

Limited functionality : The conditional operator is primarily used for simple conditional assignments. It may not be suitable for complex conditions or multiple actions, as it can quickly become unreadable and difficult to maintain.
Lack of flexibility : The conditional operator only allows for a binary choice based on the condition. It cannot handle multiple cases or multiple actions within a single line of code.
Potential for reduced readability : While the conditional operator can enhance code readability for simple assignments, it can also make the code more difficult to understand if the condition and assigned values become complex.

if-else statement

The if-else statement allows you to perform different actions based on a condition.

If the condition evaluates to true, statement 1 is executed. Otherwise, statement 2 is executed.

Read more on Verilog if-else-if statements

case statement

The case statement is used when you have multiple conditions and want to perform different actions based on the value of a variable.

The expression is evaluated, and based on its value, the corresponding statement is executed. If none of the values match the expression, the statement under default is executed.

?: conditional operator in Verilog

Compact conditional operators.

Many Verilog designs make use of a compact conditional operator:

A comman example, shown below, is an “enable” mask. Suppose there is some internal signal named a . When enabled by en== 1 , the module assigns q = a , otherwise it assigns q = 0 :

The syntax is also permitted in always blocks:

Assigned Tasks

This assignment uses only a testbench simulation, with no module to implement. Open the file src/testbench.v and examine how it is organized. It uses the conditional operator in an always block to assign q = a^b (XOR) when enabled, else q= 0 .

Run make simulate to test the operation. Verify that the console output is correct. Then modify the testbench to use an assign statement instead of an always block . Change the type of q as appropriate for the assign statement.

Turn in your work using git :

Indicate on Canvas that your assignment is done.

Verilog Conditional Operator

Just what the heck is that question mark doing.

Have you ever come across a strange looking piece of Verilog code that has a question mark in the middle of it? A question mark in the middle of a line of code looks so bizarre; they’re supposed to go at the end of sentences! However in Verilog the ? operator is a very useful one, but it does take a bit of getting used to.

The question mark is known in Verilog as a conditional operator though in other programming languages it also is referred to as a ternary operator , an inline if , or a ternary if . It is used as a short-hand way to write a conditional expression in Verilog (rather than using if/else statements). Let’s look at how it is used:

Here, condition is the check that the code is performing. This condition might be things like, “Is the value in A greater than the value in B?” or “Is A=1?”. Depending on if this condition evaluates to true, the first expression is chosen. If the condition evaluates to false, the part after the colon is chosen. I wrote an example of this. The code below is really elegant stuff. The way I look at the question mark operator is I say to myself, “Tell me about the value in r_Check. If it’s true, then return “HI THERE” if it’s false, then return “POTATO”. You can also use the conditional operator to assign signals , as shown with the signal w_Test1 in the example below. Assigning signals with the conditional operator is useful!

Nested Conditional Operators

There are examples in which it might be useful to combine two or more conditional operators in a single assignment. Consider the truth table below. The truth table shows a 2-input truth table. You need to know the value of both r_Sel[1] and r_Sel[0] to determine the value of the output w_Out. This could be achieved with a bunch of if-else if-else if combinations, or a case statement, but it’s much cleaner and simpler to use the conditional operator to achieve the same goal.

Learn Verilog

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How to use 2 condition in assign [verilog]

Thread starter daisordan
Start date Oct 17, 2012
Oct 17, 2012

Newbie level 6

always_comb //always_comb is the same function as assign begin if(a==b) z=a; else if (b==c) z=b; end Click to expand...

assign z = (a & b) ? a:z; //I know thats wrong since a=/=b , this will output z for z. What should I write if a=/=b, it will do " if (b==c) z=b; Click to expand...

Advanced Member level 3

Your always_comb blocks is not combinatorial - if a != b, and b != c, then there is no assignment to z and you have a latch. Your second if statement needs an else clause. In any case, the conditional ?: operator can be nested: assign z = (a==b) ? a : (b==c) ? b : z; That z at the end represents the missing else clause.

Full Member level 6

Advanced Member level 5

always_comb is the same function as assign Click to expand...

if(a==b) begin z=a; count=count+1; end else if (b==c) begin z=b; count=count+2; end Click to expand...

assign z = (a==b) ? (a & (count=count+1)) : (b==c) ?( b & (count=count+2)): z; Click to expand...

daisordan, If you want your code to be synthesizable, you cannot combine combinatorial and sequential logic like this in a single assign statement. You will need to explain your desired functionality without using any Verilog syntax first so we can suggest the best way to code what you want to achieve. If you don't care if your code is synthesizable, you can assign the output of a function call Code: assign z = myfunction(a,b,c); function logic myfunction(input a,b,c); if(a==b) begin z=a; count=count+1; // this is a side-effect that is not synthesizable end else if (b==c) begin z=b; count=count+2; // this is a side-effect that is not synthesizable end endfunction The biggest difference between always_comb and an assign statement is with the how the simulator deals with the semantics of function calls. An assignment statement only looks for events on the operands that appear on the RHS of the assignment, while always_comb expands functions in-line and looks for any change of any operand that appears inside the function. For example suppose I re-wrote the function to directly reference a,b,c instead of passing them as arguments to the function: Code: assign z1 = myfunction(); always_comb z2 = myfunction(); function logic myfunction(); if(a==b) z=a; else if (b==c) z=b; else z ='bx; // a don't care to prevent a latch endfunction The asssign z1= statement would never execute because there are no triggering events on the RHS to cause an evaluation. The always_comb z2= block executes at time 0, and when there is a change on any operand that is referenced within the block. mrfibble, I recommend NEVER TO USE always @(*) because it does not guarantee execution at time 0. I have Seen code like Code: real pi; `define PI 3.14159 always @(*) pi = `PI; // DO NOT EVER DO use always_comb pi = `PI; instead that fails because there was never an event to trigger the always @(*) block.

daisordan said: mrflibble: Could you briefly talk about why always_comb and assign are different? because when I read some verilog beginner's book, it said that are doing the same things. Click to expand...

dave_59 said: mrfibble, I recommend NEVER TO USE always @(*) because it does not guarantee execution at time 0. I have Seen code like Code: real pi; `define PI 3.14159 always @(*) pi = `PI; // DO NOT EVER DO use always_comb pi = `PI; instead that fails because there was never an event to trigger the always @(*) block. Click to expand...

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Using Continuous Assignment to Model Combinational Logic in Verilog

In this post, we talk about continuous assignment in verilog using the assign keyword. We then look at how we can model basic logic gates and multiplexors in verilog using continuous assignment.

There are two main classes of digital circuit which we can model in verilog – combinational and sequential .

Combinational logic is the simplest of the two, consisting solely of basic logic gates, such as ANDs, ORs and NOTs. When the circuit input changes, the output changes almost immediately (there is a small delay as signals propagate through the circuit).

In contrast, sequential circuits use a clock and require storage elements such as flip flops . As a result, output changes are synchronized to the circuit clock and are not immediate.

In this post, we talk about the techniques we can use to design combinational logic circuits in verilog. In the next post, we will discuss the techniques we use to model basic sequential circuits .

Continuous Assignment in Verilog

We use continuous assignment to drive data onto verilog net types in our designs. As a result of this, we often use continuous assignment to model combinational logic circuits.

We can actually use two different methods to implement continuous assignment in verilog.

The first of these is known as explicit continuous assignment. This is the most commonly used method for continuous assignment in verilog.

In addition, we can also use implicit continuous assignment, or net declaration assignment as it is also known. This method is less common but it can allow us to write less code.

Let's look at both of these techniques in more detail.

Explicit Continuous Assignment

We normally use the assign keyword when we want to use continuous assignment in verilog. This approach is known as explicit continuous assignment.

The verilog code below shows the general syntax for continuous assignment using the assign keyword.

The <variable> field in the code above is the name of the signal which we are assigning data to. We can only use continuous assignment to assign data to net type variables.

The <value> field can be a fixed value or we can create an expression using the verilog operators we discussed in a previous post. We can use either variable or net types in this expression.

When we use continuous assignment, the <variable> value changes whenever one of the signals in the <value> field changes state.

The code snippet below shows the most basic example of continuous assignment in verilog. In this case, whenever the b signal changes states, the value of a is updated so that it is equal to b.

Net Declaration Assignment

We can also use implicit continuous assignment in our verilog designs. This approach is also commonly known as net declaration assignment in verilog.

When we use net declaration assignment, we place a continuous assignment in the statement which declares our signal. This can allow us to reduce the amount of code we have to write.

To use net declaration assignment in verilog, we use the = symbol to assign a value to a signal when we declare it.

The code snippet below shows the general syntax we use for net declaration assignment.

The variable and value fields have the same function for both explicit continuous assignment and net declaration assignment.

As an example, the verilog code below shows how we would use net declaration assignment to assign the value of b to signal a.

Modelling Combinational Logic Circuits in Verilog

We use continuous assignment and the verilog operators to model basic combinational logic circuits in verilog.

To show we would do this, let's look at the very basic example of a three input and gate as shown below.

To model this circuit in verilog, we use the assign keyword to drive the data on to the and_out output. This means that the and_out signal must be declared as a net type variable, such as a wire.

We can then use the bit wise and operator (&) to model the behavior of the and gate.

The code snippet below shows how we would model this three input and gate in verilog.

This example shows how simple it is to design basic combinational logic circuits in verilog. If we need to change the functionality of the logic gate, we can simply use a different verilog bit wise operator .

If we need to build a more complex combinational logic circuit, it is also possible for us to use a mixture of different bit wise operators.

To demonstrate this, let's consider the basic circuit shown below as an example.

To model this circuit in verilog, we need to use a mixture of the bit wise and (&) and or (|) operators. The code snippet below shows how we would implement this circuit in verilog.

Again, this code is relatively straight forward to understand as it makes use of the verilog bit wise operators which we discussed in the last post.

However, we need to make sure that we use brackets to model more complex logic circuit. Not only does this ensure that the circuit operates properly, it also makes our code easier to read and maintain.

Modelling Multiplexors in Verilog

Multiplexors are another component which are commonly used in combinational logic circuits.

In verilog, there are a number of ways we can model these components.

One of these methods uses a construct known as an always block . We normally use this construct to model sequential logic circuits, which is the topic of the next post in this series. Therefore, we will look at this approach in more detail the next blog post.

In the rest of this post, we will look at the other methods we can use to model multiplexors.

Verilog Conditional Operator

As we talked about in a previous blog, there is a conditional operator in verilog . This functions in the same way as the conditional operator in the C programming language.

To use the conditional operator, we write a logical expression before the ? operator which is then evaluated to see if it is true or false.

The output is assigned to one of two values depending on whether the expression is true or false.

The verilog code below shows the general syntax which the conditional operator uses.

From this example, it is clear how we can create a basic two to one multiplexor using this operator.

However, let's look at the example of a simple 2 to 1 multiplexor as shown in the circuit diagram below.

The code snippet below shows how we would use the conditional operator to model this multiplexor in verilog.

Nested Conditional Operators

Although this is not common, we can also write code to build larger multiplexors by nesting conditional operators.

To show how this is done, let's consider a basic 4 to 1 multiplexor as shown in the circuit below.

To model this in verilog using the conditional operator, we treat the multiplexor circuit as if it were a pair of two input multiplexors.

This means one multiplexor will select between inputs A and B whilst the other selects between C and D. Both of these multiplexors use the LSB of the address signal as the address pin.

To create the full four input multiplexor, we would then need another multiplexor.

This takes the outputs from the first two multiplexors and uses the MSB of the address signal to select between them.

The code snippet below shows the simplest way to do this. This code uses the signals mux1 and mux2 which we defined in the last example.

However, we could easily remove the mux1 and mux2 signals from this code and instead use nested conditional operators.

This reduces the amount of code that we would have to write without affecting the functionality.

The code snippet below shows how we would do this.

As we can see from this example, when we use conditional operators to model multiplexors in verilog, the code can quickly become difficult to understand. Therefore, we should only use this method to model small multiplexors.

Arrays as Multiplexors

It is also possible for us to use verilog arrays to build simple multiplexors.

To do this we combine all of the multiplexor inputs into a single array type and use the address to point at an element in the array.

To get a better idea of how this works in practise, let's consider a basic four to one multiplexor as an example.

The first thing we must do is combine our input signals into an array. There are two ways in which we can do this.

Firstly, we can declare an array and then assign all of the individual bits, as shown in the verilog code below.

Alternatively we can use the verilog concatenation operator , which allows us to assign the entire array in one line of code.

To do this, we use a pair of curly braces - { } - and list the elements we wish to include in the array inside of them.

When we use the concatenation operator we can also declare and assign the variable in one statement, as long as we use a net type.

The verilog code below shows how we can use the concatenation operator to populate an array.

As verilog is a loosely typed language , we can use the two bit addr signal as if it were an integer type. This signal then acts as a pointer that determines which of the four elements to select.

The code snippet below demonstrates this method in practise. As the mux output is a wire, we must use continuous assignment in this instance.

What is the difference between implicit and explicit continuous assignment?

When we use implicit continuous assignment we assign the variable a value when we declare. When we use explicit continuous assignment we use the assign keyword to assign a value.

Write the code for a 2 to 1 multiplexor using any of the methods discussed we discussed.

Write the code for circuit below using both implicit and explicit continuous assignment.