alu

问题 I'm trying to test if a wire(s) is on or not to signify if there is an error/overflow in my alu code. Given this code: output reg[3:0]x; // line 149 output wire error; output wire overflow; always @* begin if(error || overflow) begin assign x = 4'b1111; // line 155 assign error = ~error; assign overflow = ~overflow; end else begin assign x = opcode; end end I get following error messages: uut is my instantiation unit in my testbench called main 回答1: The code in the example has several issues.

问题 I'm trying to test if a wire(s) is on or not to signify if there is an error/overflow in my alu code. Given this code: output reg[3:0]x; // line 149 output wire error; output wire overflow; always @* begin if(error || overflow) begin assign x = 4'b1111; // line 155 assign error = ~error; assign overflow = ~overflow; end else begin assign x = opcode; end end I get following error messages: uut is my instantiation unit in my testbench called main 回答1: The code in the example has several issues.

问题 I was watching some lecture on algorithms, and the professor used multiplication as an example of how naive algorithms can be improved... It made me realize that multiplication is not that obvious, although when I am coding I just consider it a simple atomic operation, multiplication requires a algorithm to run, it does not work like summing numbers. So I wonder, what algorithm modern desktop processors actually use? I guess they don't rely on logarithm tables, and don't make loops with

问题 I was watching some lecture on algorithms, and the professor used multiplication as an example of how naive algorithms can be improved... It made me realize that multiplication is not that obvious, although when I am coding I just consider it a simple atomic operation, multiplication requires a algorithm to run, it does not work like summing numbers. So I wonder, what algorithm modern desktop processors actually use? I guess they don't rely on logarithm tables, and don't make loops with

问题 Closed . This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed last year . My code works if I use operand 1 and operand 2 as integers. Using unsigned char operand 1 does not work. Can you help me? int ALU(unsigned char operand1, unsigned char operand2) { printf("Enter Operand 1(in decimal): "); scanf("%d",&operand1); printf("\nEnter Operand 2(in decimal): "); scanf("%d",

问题 I believe integer addition or subtraction always take the same time no matter how big the operands are. Time needed for ALU output to be stabilized may vary over input operands, but CPU component that exploits ALU output will wait sufficiently long time so that any integer operation will be processed in SAME cycles. (Cycles needed for ADD, SUB, MUL, and DIV will be different, but ADD will take the same cycles regardless of input operands, I think.) Is this true for floating point operation,

问题 just a simple question regarding calculations in IJVM as I couldn't find the solution in their documentation. Suppose we need to perform the following calculation: BIPUSH 0x32 // PUSH 2 BIPUSH 0x33 // PUSH 3 IADD // PUSH sum(2,3) OUT // output: "e" IADD ----> Pop two words from stack; push their sum I know the solution is likely straight forward, but for the life of me I can't recall on how to convert the addition/output to the actual digits. How to make it output "5" instead of this stupid

问题 -This code is written in verilog using Modelsim 10.2d.The errors below indicate there is some problem with {cout,l3} assignment. module alu(a,b,bin,cin,op,cout,res); input [31:0] a,b; input [1:0] op; input bin,cin; reg [31:0] l1,l2,l3; output cout; output [31:0] res; assign l1 = a & b; assign l2 = a | b; initial if(bin == 1'b0) assign {cout,l3} = a + b + cin; else assign {cout,l3} = a - b + cin; mux4to1(l1,l2,l3,op,res); endmodule Error- v(14): LHS in procedural continuous assignment may not

问题 Here is the sheet that I need to follow https://imgur.com/a/JuLpQZt Here is my code as of now #include<stdio.h> #include<conio.h> #include<stdlib.h> #include<windows.h> void arithmetic(); int ALU(unsigned char operand1, unsigned char operand2, unsigned char control_signals); int askOperand1(); int askOperand2(); unsigned char operand1; unsigned char operand2; unsigned char control_signals; const int ACC = 16; //ACC = ACCUMULATOR int main() { for(;;) { system("cls"); ALU(operand2,operand2

问题 I have basic binary math question. For example; reg [31:0] a = 32'hFFFF_FFFF; reg [31:0] b = 32'hFFFF_FFFF; reg [31:0] c = 0; I know c = a - b will result in zero and overflow flag will not be set. My understanding is that ALU uses 2's complement of subtrahend (right side of operator) and adds it to minuend (left side of operator). So 2s complement for value stored in reg b is 32'h1. Now, if I add this to reg a, I will get 32 zeros and a 1, which is overflow. Then why is not overflow flag set