2016-11-11 22 views
0

我有一點問題。我想做一個用固定點來分兩個數字的程序。我的輸出看起來不錯,但是當我在 本地測試網站上測試程序時,我得到了一些奇怪的結果。有人可以幫助我嗎?謝謝。定點除數

module divider(
     input[7:0] a,b, 
     output reg [15:0] q, 
     output reg [7:0] r,frac 
     ); 

     reg[7:0] c2, r2; 
     integer c; 
    always @(*) 
     begin 

     c = 8'b00000000; 
     r = a; 
      frac = 8'b00000000; 

     repeat (30) 
      begin 
       if (r >= b) 
        begin 
        c = c + 1; 
        r = r - b; 
        end 
      end 

      q[15:8] = c; 



     repeat (8) 
     begin 
      if(r != 0) 
       begin 
       r = r*10; 
       c2 = 8'b00000000; 

       repeat(30) 
        begin 

        if (r>=b) 
         begin 
         c2 = c2 + 1; 
         r = r-b; 
         end 
        end 
       frac = frac*10 + c2; 
       end 
      end 

      q[7:0] = frac; 








end 

endmodule 

輸出本地站點:

error: a = 1, b = 2, q = 0.019531, expected q = 0.500000 
error: a = 1, b = 4, q = 0.097656, expected q = 0.250000 
error: a = 1, b = 5, q = 0.007813, expected q = 0.199219 
error: a = 1, b = 7, q = 0.570313, expected q = 0.140625 
error: a = 1, b = 8, q = 0.488281, expected q = 0.125000 
error: a = 1, b = 9, q = 0.777344, expected q = 0.109375 
error: a = 1, b = 10, q = 0.003906, expected q = 0.097656 
error: a = 1, b = 11, q = 0.363281, expected q = 0.089844 
error: a = 1, b = 14, q = 0.785156, expected q = 0.070313 
error: a = 1, b = 15, q = 0.664063, expected q = 0.066406 
error: a = 1, b = 16, q = 0.441406, expected q = 0.062500 
error: a = 1, b = 17, q = 0.937500, expected q = 0.058594 
error: a = 1, b = 18, q = 0.386719, expected q = 0.054688 
error: a = 1, b = 19, q = 0.207031, expected q = 0.050781 
error: a = 1, b = 20, q = 0.019531, expected q = 0.046875 
error: a = 1, b = 21, q = 0.187500, expected q = 0.046875 
error: a = 1, b = 22, q = 0.679688, expected q = 0.042969 
error: a = 1, b = 23, q = 0.695313, expected q = 0.042969 
error: a = 1, b = 25, q = 0.015625, expected q = 0.039063 
error: a = 1, b = 27, q = 0.589844, expected q = 0.035156 
error: a = 1, b = 28, q = 0.890625, expected q = 0.035156 
error: a = 1, b = 29, q = 0.824219, expected q = 0.031250 
error: a = 1, b = 30, q = 0.832031, expected q = 0.031250 
error: a = 1, b = 31, q = 0.804688, expected q = 0.031250 
error: a = 1, b = 32, q = 0.207031, expected q = 0.031250 
error: a = 1, b = 33, q = 0.121094, expected q = 0.027344 
error: a = 1, b = 34, q = 0.437500, expected q = 0.027344 
error: a = 1, b = 35, q = 0.570313, expected q = 0.027344 
error: a = 1, b = 36, q = 0.914063, expected q = 0.027344 
error: a = 1, b = 37, q = 0.000000, expected q = 0.023438 
error: a = 1, b = 38, q = 0.339844, expected q = 0.023438 
error: a = 1, b = 40, q = 0.097656, expected q = 0.023438 
+0

不是100%確定該算法是如何支持工作的;但是如果你通過你的第一個例子,你會發現生成的'q'從你正在做的事情中是有意義的。請注意,這種設計描述了完全可能是或不是您的意圖的組合邏輯。我不知道你從哪裏得到'repeat(30)',從例子中你可以看到'a = 8'd100','b = 8'd2','c'在第一個循環結束時會是30將不會做足夠的連續減法。另外請注意,固定點通常是基數2,而不是基數10,因爲您在這裏('fact = fact * 10 + c2')。 – Unn

回答

0

你不一定需要基數10對FPGA劃分。 Radix2部門(例如shift and subtract)將更快,更容易實施。我發現this爲radix2劃分的例子。