2.4、Procedures
目录
2.4.1、Always blocks(combinational)
2.4.7、Priority encoder with casez:
2.4.1、Always blocks(combinational)
problem statement:
Since digital circuits are composed of logic gates connected with wires, any circuit can be expressed as some combination of modules and assign statements. However, sometimes this is not the most convenient way to describe the circuit. Procedures (of which always blocks are one example) provide an alternative syntax for describing circuits.
For synthesizing hardware, two types of always blocks are relevant:
- Combinational: always @(*)
- Clocked: always @(posedge clk)
Build an AND gate using both an assign statement and a combinational always block. (Since assign statements and combinational always blocks function identically, there is no way to enforce that you're using both methods. But you're here for practice, right?...)
solution:
// synthesis verilog_input_version verilog_2001
module top_module(
input a,
input b,
output wire out_assign,
output reg out_alwaysblock
);
assign out_assign=a&b;
always @(*) begin
if(a==1&&b==1)
out_alwaysblock<=1;
else
out_alwaysblock<=0;
end
endmodule注：always语句中使用非阻塞赋值。
2.4.2、always blocks （clocked）
problem statement:
Build an XOR gate three ways, using an assign statement, a combinational always block, and a clocked always block. Note that the clocked always block produces a different circuit from the other two: There is a flip-flop so the output is delayed.
solution:
// synthesis verilog_input_version verilog_2001
module top_module(
input clk,
input a,
input b,
output wire out_assign,
output reg out_always_comb,
output reg out_always_ff );
assign out_assign=a^b;
always@(*) begin
out_always_comb=a^b;
end
always@(posedge clk) begin
out_always_ff<=a^b;
end
endmodule注：该题为我们展示了连续赋值语句 过程赋值语句中的非阻塞赋值和阻塞赋值的区别 。
2.4.3 if statement
problem statement：
Build a 2-to-1 mux that chooses between a and b. Choose b if both sel_b1 and sel_b2 are true. Otherwise, choose a. Do the same twice, once using assign statements and once using a procedural if statement.
solution：
// synthesis verilog_input_version verilog_2001
module top_module(
input a,
input b,
input sel_b1,
input sel_b2,
output wire out_assign,
output reg out_always );
assign out_assign=(sel_b1==1&&sel_b2==1)?b:a;
always@(*) begin
if(sel_b1==1&&sel_b2==1)
begin out_always=b;
end
else begin
out_always=a;
end
end
endmodule注意assign赋值语句后的问号表达式为真时，则out_assign=b否则out_assign=a
2.4.4、 if statement latches
problem statement:
solution:
// synthesis verilog_input_version verilog_2001
module top_module (
input cpu_overheated,
output reg shut_off_computer,
input arrived,
input gas_tank_empty,
output reg keep_driving ); //
always @(*) begin
if (cpu_overheated)
shut_off_computer = 1;
else
shut_off_computer=0;
end
always @(*) begin
if (~arrived)
keep_driving = ~gas_tank_empty;
else
keep_driving=0;
end
endmodule缺少else
2.4.5、Case statement:
problem statement:
Case statements are more convenient than if statements if there are a large number of cases. So, in this exercise, create a 6-to-1 multiplexer. When sel is between 0 and 5, choose the corresponding data input. Otherwise, output 0. The data inputs and outputs are all 4 bits wide.
Be careful of inferring latches (See.always_if2)
solution:
// synthesis verilog_input_version verilog_2001
module top_module (
input [2:0] sel,
input [3:0] data0,
input [3:0] data1,
input [3:0] data2,
input [3:0] data3,
input [3:0] data4,
input [3:0] data5,
output reg [3:0] out );//
always@(*) begin // This is a combinational circuit
case(sel)
0:out=data0;
1:out=data1;
2:out=data2;
3:out=data3;
4:out=data4;
5:out=data5;
default:out=0;
endcase
end
endmodule注：case的语法练习，注意不要丢掉endcase
2.4. 6、Priority encoder
problem statement:
A priority encoder is a combinational circuit that, when given an input bit vector, outputs the position of the first 1 bit in the vector. For example, a 8-bit priority encoder given the input 8'b10010000 would output 3'd4, because bit[4] is first bit that is high.
Build a 4-bit priority encoder. For this problem, if none of the input bits are high (i.e., input is zero), output zero. Note that a 4-bit number has 16 possible combinations.
Using hexadecimal (4'hb) or decimal (4'd11) number literals would save typing vs. binary (4'b1011) literals.
solution:
// synthesis verilog_input_version verilog_2001
module top_module (
input [3:0] in,
output reg [1:0] pos );
always @(*) begin
casez (in[3:0])
4'bzzz1: pos = 0; // in[3:1] can be anything
4'bzz10: pos = 1;
4'bz100: pos = 2;
4'b1000: pos = 3;
default: pos = 0;
endcase
end
endmodule2.4.7、Priority encoder with casez:
problem statement:
Build a priority encoder for 8-bit inputs. Given an 8-bit vector, the output should report the first bit in the vector that is 1. Report zero if the input vector has no bits that are high. For example, the input 8'b10010000 should output 3'd4, because bit[4] is first bit that is high.
From the previous exercise (always_case2), there would be 256 cases in the case statement. We can reduce this (down to 9 cases) if the case items in the case statement supported don't-care bits. This is what casez is for: It treats bits that have the value z as don't-care in the comparison.
A case statement behaves as though each item is checked sequentially (in reality, it does something more like generating a giant truth table then making gates). Notice how there are certain inputs (e.g., 4'b1111) that will match more than one case item. The first match is chosen (so 4'b1111 matches the first item, out = 0, but not any of the later ones).
- There is also a similar casex that treats both x and z as don't-care. I don't see much purpose to using it over casez.
- The digit ? is a synonym for z. so 2'bz0 is the same as 2'b?0
- Binary literals have to be used here because you need to specify z for some bits.
solution:
// synthesis verilog_input_version verilog_2001
module top_module (
input [7:0] in,
output reg [2:0] pos );
always@(*)
casez(in)
8'bzzzzzzz1:pos=0;
8'bzzzzzz10:pos=1;
8'bzzzzz100:pos=2;
8'bzzzz1000:pos=3;
8'bzzz10000:pos=4;
8'bzz100000:pos=5;
8'bz1000000:pos=6;
8'b10000000:pos=7;
default:pos=0;
endcase
endmodule注意关键字是casez不是case
2.4.8、Avoiding latches:
problem statement:
Suppose you're building a circuit to process scancodes from a PS/2 keyboard for a game. Given the last two bytes of scancodes received, you need to indicate whether one of the arrow keys on the keyboard have been pressed. This involves a fairly simple mapping, which can be implemented as a case statement (or if-elseif) with four cases.
Your circuit has one 16-bit input, and four outputs. Build this circuit that recognizes these four scancodes and asserts the correct output.
To avoid creating latches, all outputs must be assigned a value in all possible conditions (See also always_if2). Simply having a default case is not enough. You must assign a value to all four outputs in all four cases and the default case. This can involve a lot of unnecessary typing. One easy way around this is to assign a "default value" to the outputs before the case statement:
solution:
// synthesis verilog_input_version verilog_2001
module top_module (
input [15:0] scancode,
output reg left,
output reg down,
output reg right,
output reg up );
always@(*) begin
up=1'b0;down=1'b0;left=1'b0;right=1'b0;
case(scancode)
16'he06b:left=1'b1;
16'he072:down=1'b1;
16'he074:right=1'b1;
16'he075:up=1'b1;
default:up=1'b0;
endcase
end
endmodule
可以用十六进制表示scancode