256个组,每个组4个cache行,用lru替换算法实现替换 verilog实现
时间: 2024-02-27 14:53:04 浏览: 141
下面是一个简单的 Verilog 实现,其中使用了一个 LRU 计数器来确定要替换的缓存行:
```
module Cache (
input [31:0] addr,
input [31:0] data_in,
input wr_en,
output [31:0] data_out
);
parameter NUM_SETS = 256;
parameter NUM_LINES = 4;
parameter NUM_WORDS = 8;
parameter TAG_WIDTH = 20;
reg [TAG_WIDTH-1:0] tag [NUM_SETS-1:0][NUM_LINES-1:0];
reg [NUM_WORDS*8-1:0] data [NUM_SETS-1:0][NUM_LINES-1:0];
reg [NUM_LINES-1:0] lru [NUM_SETS-1:0];
wire [NUM_WORDS*8-1:0] data_out;
assign data_out = data[addr[NUM_SETS+1:3]][lru[addr[NUM_SETS+1:3]]];
always @ (posedge clk) begin
if (wr_en) begin
int set_index = addr[NUM_SETS+1:3];
int line_index = 0;
int lru_line = 0;
int lru_count = 0;
for (int i = 0; i < NUM_LINES; i = i + 1) begin
if (tag[set_index][i] == addr[TAG_WIDTH+NUM_SETS+2:NUM_SETS+3]) begin
line_index = i;
break;
end
if (lru[set_index][i] > lru_count) begin
lru_count = lru[set_index][i];
lru_line = i;
end
end
if (line_index != 0) begin
tag[set_index][line_index] = addr[TAG_WIDTH+NUM_SETS+2:NUM_SETS+3];
data[set_index][line_index] = data_in;
lru[set_index][line_index] = 0;
for (int i = 0; i < NUM_LINES; i = i + 1) begin
if (i != line_index) begin
lru[set_index][i] = lru[set_index][i] + 1;
end
end
end else begin
tag[set_index][lru_line] = addr[TAG_WIDTH+NUM_SETS+2:NUM_SETS+3];
data[set_index][lru_line] = data_in;
lru[set_index][lru_line] = 0;
for (int i = 0; i < NUM_LINES; i = i + 1) begin
if (i != lru_line) begin
lru[set_index][i] = lru[set_index][i] + 1;
end
end
end
end
end
endmodule
```
这个实现为每个组维护了一个 LRU 计数器,用于确定要替换的缓存行。当写入数据时,它会查找与地址匹配的缓存行,如果找到则更新缓存行的标记、数据和 LRU 计数器。如果没有找到匹配的缓存行,则使用 LRU 计数器确定要替换的缓存行,并将新数据写入该行。注意,这个实现是为了说明 LRU 替换算法而设计的,它可能需要根据实际需求进行修改。
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