Math  /  Discrete

QuestionQuestion 2 (3 Points) Print the output of the following Verilog code ``` module tb(); int a,b,c,d,e; initial begin #100; display("a=display("a = %0d",a); display("b = %0d",b); display("c=display("c = %Od",c); display("d = %0d",d); $display("e = %Od",e); end ``` initial begin a='b011; b='b111; c='b100; d='b101; e='b001; a = b + c + d; b <= a + d + e; c = a + b + e; d <= b + c + e; e = c + d + a; a <= d + e + b; b = a + c + d; end endmodule

Studdy Solution

STEP 1

1. The Verilog code provided is a testbench module.
2. The variables a,b,c,d,e a, b, c, d, e are declared as integers.
3. The initial block contains both blocking (`=`) and non-blocking (`<=`) assignments.
4. The code execution is based on the order of assignments and delays.
5. The code will print the values of a,b,c,d,e a, b, c, d, e after a delay of 100 time units.

STEP 2

1. Analyze the initial assignments and their effects.
2. Determine the sequence of operations and how they affect the variables.
3. Calculate the final values of the variables at the time of display.

STEP 3

Initially, the variables are assigned binary values: - a=b011 a = 'b011 which is 3 in decimal. - b=b111 b = 'b111 which is 7 in decimal. - c=b100 c = 'b100 which is 4 in decimal. - d=b101 d = 'b101 which is 5 in decimal. - e=b001 e = 'b001 which is 1 in decimal.

STEP 4

Evaluate the blocking assignments: - a=b+c+d=7+4+5=16 a = b + c + d = 7 + 4 + 5 = 16 - c=a+b+e=16+7+1=24 c = a + b + e = 16 + 7 + 1 = 24 - e=c+d+a=24+5+16=45 e = c + d + a = 24 + 5 + 16 = 45 - b=a+c+d=16+24+5=45 b = a + c + d = 16 + 24 + 5 = 45

STEP 5

Evaluate the non-blocking assignments: - b b is assigned non-blocking, but since it is re-assigned with blocking, the non-blocking assignment is overridden. - d<=b+c+e=7+4+1=12 d <= b + c + e = 7 + 4 + 1 = 12 (using initial values of b,c,e b, c, e ) - a<=d+e+b=5+1+7=13 a <= d + e + b = 5 + 1 + 7 = 13 (using initial values of d,e,b d, e, b )

STEP 6

After 100 time units, the following values are printed: - a=13 a = 13 - b=45 b = 45 - c=24 c = 24 - d=12 d = 12 - e=45 e = 45
The output of the Verilog code is:
``` a = 13 b = 45 c = 24 d = 12 e = 45 ```

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