verilog 实现32位浮点数减法器
时间: 2023-10-22 18:06:43 浏览: 53
以下是一个简单的32位浮点数减法器的Verilog实现:
```verilog
module float_subtractor(
input [31:0] a,
input [31:0] b,
output reg [31:0] result
);
reg [30:0] mantissa_a;
reg [30:0] mantissa_b;
reg [8:0] exponent_a;
reg [8:0] exponent_b;
reg sign_a;
reg sign_b;
assign sign_a = a[31];
assign sign_b = b[31];
assign mantissa_a = {1'b1, a[22:0]};
assign mantissa_b = {1'b1, b[22:0]};
assign exponent_a = a[30:23];
assign exponent_b = b[30:23];
always @(*) begin
if (exponent_a > exponent_b) begin
mantissa_b = mantissa_b >> (exponent_a - exponent_b);
exponent_b = exponent_a;
end else if (exponent_b > exponent_a) begin
mantissa_a = mantissa_a >> (exponent_b - exponent_a);
exponent_a = exponent_b;
end
end
always @(*) begin
if (sign_a && !sign_b) begin
mantissa_a = ~mantissa_a + 1;
result[31] = 1'b1;
end else if (!sign_a && sign_b) begin
mantissa_b = ~mantissa_b + 1;
end
if (mantissa_a >= mantissa_b) begin
result[22:0] = mantissa_a - mantissa_b;
end else begin
result[22:0] = mantissa_b - mantissa_a;
result[31] = ~result[31];
end
result[30:23] = exponent_a;
result[31] = sign_a;
end
endmodule
```
该减法器接收两个32位浮点数A和B,计算A-B并将结果存储在32位寄存器中。该代码首先从输入中提取A和B的符号、尾数和指数。然后,它将尾数和指数对齐(如果必要),并执行减法操作。最后,它重新组合符号、尾数和指数以生成结果。