If A=1’b1,B=2’b01,C=2’b00 y={4{A},2{B},C} equalsa.Noneb.10’b1111010100c. 8’b11110100d.9’b111101010
Question
If A=1’b1,B=2’b01,C=2’b00 y={4{A},2{B},C} equalsa.Noneb.10’b1111010100c. 8’b11110100d.9’b111101010
Solution
The given values are in binary and Verilog format.
- A is 1'b1 which means it is a 1-bit binary number with a value of 1.
- B is 2'b01 which means it is a 2-bit binary number with a value of 01.
- C is 2'b00 which means it is a 2-bit binary number with a value of 00.
The expression y={4{A},2{B},C} is a concatenation operation in Verilog.
- 4{A} means repeating A four times, which gives 1111.
- 2{B} means repeating B two times, which gives 0101.
- C is 00.
Concatenating these values gives 1111010100.
So, the answer is b. 10'b1111010100.
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