我尝试使用define来简化编写,所有的参数都是真正编写的
//...
parameter BLEZ = 1001011;
parameter BLTZ = 1001100;
parameter SRA = 1001101;
`define R_type1 ((op == MOVA)||(op == MOVB)||(op == ADD)||(op == SUB)||(op == AND)||(op == OR)||(op == XOR)||(op == NOT)||(op == SLT))
`define R_type2 ((op == LSL)||(op == LSR)||(op == SRA))
`define JR_type ((op == JMR))
`define J_type ((op == JMP))
`define I_type ((op == ADI)||(op == SBI)||(op == ANI)||(op == ORI)||(op == XRI)||(op == AIU)||(op == SIU)(op == JML))
`define LW ((op == LD))
`define SW ((op == ST))
`define Branch ((op == BZ)||(op == BNZ)||(op == BGEZ)||(op == BGTZ)||(op == BLEZ)||(op == BLTZ))
always@(op)
begin
RegWrite_id = ((`LW)||(`R_type1)||(`R_type2)||(`I_type));
RegDst_id = ((`LW)||(`R_type1)||(`R_type2)||(`I_type));
MemWrite_id = 1;//(`SW);
MemRead_id = (`LW);
MemToReg_id = (`LW);
ALUSrcA_id = (`R_type2);
ALUSrcB_id = (`I_type);
PCSource = {`JR_type,`J_type,Z};
end
我认为在逻辑层面上没有什么问题,但它总是会给出这样的错误:
RegWrite_id = ((`LW)||(`R_type1)||(`R_type2)||(`I_type));
|
ncvlog: *E,EXPRPA (Decode_Unit.v,71|56): expecting a right parenthesis (')') [4.3][9.7(IEEE)].
(`define macro: I_type [Decode_Unit.v line 64], file: Decode_Unit.v line 71)
RegWrite_id = ((`LW)||(`R_type1)||(`R_type2)||(`I_type));
|
ncvlog: *E,EXPSMC (Decode_Unit.v,71|58): expecting a semicolon (';') [9.2.2(IEEE)].
RegWrite_id = ((`LW)||(`R_type1)||(`R_type2)||(`I_type));
|
ncvlog: *E,NOTSTT (Decode_Unit.v,71|58): expecting a statement [9(IEEE)].
RegDst_id = ((`LW)||(`R_type1)||(`R_type2)||(`I_type));
|
ncvlog: *E,EXPLPA (Decode_Unit.v,72|15): expecting a left parenthesis ('(') [12.1.2][7.1(IEEE)].
RegDst_id = ((`LW)||(`R_type1)||(`R_type2)||(`I_type));
|
ncvlog: *E,EXPRPA (Decode_Unit.v,72|56): expecting a right parenthesis (')') [4.3][9.7(IEEE)].
(`define macro: I_type [Decode_Unit.v line 64], file: Decode_Unit.v line 72)
RegDst_id = ((`LW)||(`R_type1)||(`R_type2)||(`I_type));
|
ncvlog: *E,EXPSMC (Decode_Unit.v,72|58): expecting a semicolon (';') [12.1.2][7.1(IEEE)].
MemWrite_id = 1;//(`SW);
|
ncvlog: *E,EXPLPA (Decode_Unit.v,73|15): expecting a left parenthesis ('(') [12.1.2][7.1(IEEE)].
MemRead_id = (`LW);
所有的参数都是真正写好的
这让人困惑..。
请给我一些指导
发布于 2016-12-30 11:10:07
您忘记了I_type
宏中最后两个项之间的||
运算符。
另请注意,如果您希望将参数解释为二进制数,则必须在参数前面添加'b
,例如,'b1010
是二进制数字10,而1010
是1,010。
发布于 2017-01-04 03:57:23
正如Unn已经指出的,I_type
缺少||
和,需要指定基数(否则假定为十进制)。
示例
'b1010
是二进制数字10,而1010
是1,010。
此外,正如您所经历的那样,`define
的调试是一个痛苦的过程。请注意,`define
根据编译顺序应用于全局空间。这意味着另一个模块可以使用相同的宏,而不定义它,如果它是稍后编译的,这可能会导致混乱和错误不能正确完成。建议避免(或至少最小化)在RTL中使用`define
。
对于您的代码,我建议将您的`define
更改为wire
。它将合成到相同的,但更容易调试。为了演示,我明智地留下了||
的bug。
wire R_type1 = ((op == MOVA)||(op == MOVB)||(op == ADD)||(op == SUB)||(op == AND)||(op == OR)||(op == XOR)||(op == NOT)||(op == SLT));
wire R_type2 = ((op == LSL)||(op == LSR)||(op == SRA));
wire JR_type = ((op == JMR));
wire J_type = ((op == JMP));
wire I_type = ((op == ADI)||(op == SBI)||(op == ANI)||(op == ORI)||(op == XRI)||(op == AIU)||(op == SIU)(op == JML));
wire LW = ((op == LD));
wire SW = ((op == ST));
wire Branch = ((op == BZ)||(op == BNZ)||(op == BGEZ)||(op == BGTZ)||(op == BLEZ)||(op == BLTZ));
always@* // IMPORTANT :: use '*', not 'op'
begin
RegWrite_id = ((LW)||(R_type1)||(R_type2)||(I_type));
RegDst_id = ((LW)||(R_type1)||(R_type2)||(I_type));
MemWrite_id = 1;//(SW);
MemRead_id = (LW);
MemToReg_id = (LW);
ALUSrcA_id = (R_type2);
ALUSrcB_id = (I_type);
PCSource = {JR_type,J_type,Z};
end
组合always块应该使用自动敏感;always@*
或同义的always@(*)
。自IEEE1364-2001以来一直支持自动灵敏度。指定敏感度列表仅建议您严格遵循IEEE1364-1995,或者用于不会综合的行为模块中的排除信号。
https://stackoverflow.com/questions/41354897
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