在Verilog中使用一个整数生成块

Question

我试图跟踪程序在Verilog中通过嵌套的生成块运行的次数，并将其用作数组的索引，但它似乎需要一个常量该指数。有没有办法做到这一点？ //我只是通过手动创建生成模块测试，它的工作原理，但我还是想知道，如果我试图做的是可能这里是我的代码

module fivbmulti(input [4:0] xIn, yIn, 
        output [9:0] pOut); 

wire [24:0] passSum, passCout, a_b; 
integer count = 0;   //This is what I would like to use for the index 
generate 
genvar i; 
for (i = 0; i <= 4; i = i+1) begin: row 
    genvar j; 
     for (j = 0; j <=4; j = j + 1) begin: col 
      assign a_b[count] = xIn[j] & yIn[i]; //use it here as the index for a_b 
      count = count + 1;  //increment it here after each generation 
      end 
    end 
endgenerate 

assign pOut[0] = a_b[0]; 

adder row0col0(.aIn(a_b[1]), .bIn(a_b[5]), .cIn(1'b0), .cOut(passCout[0]), .sum(pOut[1])); 
adder row1col1(.aIn(a_b[2]), .bIn(a_b[6]), .cIn(1'b0), .cOut(passCout[1]), .sum(passSum[1])); 
adder row1col2(.aIn(a_b[3]), .bIn(a_b[7]), .cIn(1'b0), .cOut(passCout[2]), .sum(passSum[2])); 
adder row1col3(.aIn(a_b[4]), .bIn(a_b[8]), .cIn(1'b0), .cOut(passCout[3]), .sum(passSum[3])); 

adder row2col0(.aIn(passSum[1]), .bIn(a_b[10]), .cIn(passCout[0]), .cOut(passCout[4]), .sum(pOut[2])); 
adder row2col1(.aIn(passSum[2]), .bIn(a_b[11]), .cIn(passCout[1]), .cOut(passCout[5]), .sum(passSum[5])); 
adder row2col2(.aIn(passSum[3]), .bIn(a_b[12]), .cIn(passCout[2]), .cOut(passCout[6]), .sum(passSum[6])); 
adder row2col3(.aIn(a_b[9]), .bIn(a_b[13]), .cIn(passCout[3]), .cOut(passCout[7]), .sum(passSum[7])); 

adder row3col0(.aIn(passSum[5]), .bIn(a_b[15]), .cIn(passCout[4]), .cOut(passCout[8]), .sum(pOut[3])); 
adder row3col1(.aIn(passSum[6]), .bIn(a_b[16]), .cIn(passCout[5]), .cOut(passCout[9]), .sum(passSum[9])); 
adder row3col2(.aIn(passSum[7]), .bIn(a_b[17]), .cIn(passCout[6]), .cOut(passCout[10]), .sum(passSum[10])); 
adder row3col3(.aIn(a_b[14]), .bIn(a_b[18]), .cIn(passCout[7]), .cOut(passCout[11]), .sum(passSum[11])); 

adder row4col0(.aIn(passSum[9]), .bIn(a_b[20]), .cIn(passCout[8]), .cOut(passCout[12]), .sum(pOut[4])); 
adder row4col1(.aIn(passSum[10]), .bIn(a_b[21]), .cIn(passCout[9]), .cOut(passCout[13]), .sum(passSum[13])); 
adder row4col2(.aIn(passSum[11]), .bIn(a_b[22]), .cIn(passCout[10]), .cOut(passCout[14]), .sum(passSum[14])); 
adder row4col3(.aIn(a_b[19]), .bIn(a_b[23]), .cIn(passCout[11]), .cOut(passCout[15]), .sum(passSum[15])); 

adder row5col0(.aIn(passSum[13]), .bIn(1'b0), .cIn(passCout[12]), .cOut(passCout[16]), .sum(pOut[5])); 
adder row5col1(.aIn(passSum[14]), .bIn(passCout[16]), .cIn(passCout[13]), .cOut(passCout[17]), .sum(pOut[6])); 
adder row5col2(.aIn(passSum[15]), .bIn(passCout[17]), .cIn(passCout[14]), .cOut(passCout[18]), .sum(pOut[7])); 
adder row5col3(.aIn(a_b[24]), .bIn(passCout[18]), .cIn(passCout[15]), .cOut(pOut[9]), .sum(pOut[8])); 



endmodule 

Answer 1

如果语法的那种风格有可能计数必须是一个genvar，但它似乎并不像你那样使用它。你应该完全基于它i和j像i*4+j：

generate 
genvar i; 
genvar j; 
    for (i = 0; i <= 4; i = i+1) begin: row 
    for (j = 0; j <=4; j = j + 1) begin: col 
     assign a_b[i*4+j] = xIn[j] & yIn[i]; //use it here as the index for a_b 
    end 
    end 
endgenerate 

我已经使用i * (number of iterations of j)+ j，应计算为计数递增等同。