Question

verilog错误：cannot match operand(s) in the condition

这是做一个led自动256次变色的程序，但现在不知道哪里错了，求解。

module pwm_test6(clk,reset,led);
input clk;//50Mhz
input reset;

output led;
reg[32:0] count;
reg [25:0] timecounter;
reg [7:0] pwm_count;
reg flag;
reg pwm_flag;

always@(posedge clk or negedge reset)
begin
if(!reset)
timecounter = 0;
else
timecounter = timecounter +26'd1;

if(timecounter == 26'd49999999)
timecounter = 0;
end

always@(posedge clk)
begin
count = count+1'b1;
if(count[11:4]<pwm_count)
pwm_flag = 1;
else
pwm_flag = 0;
if (count[15] == 1'b1)
begin
if(flag == 1'b1)
begin
flag = 1'b0;
if(timecounter == 26'd25000000)
pwm_count = (pwm_count + 8'b00000001);

end
end
else
flag = 1'b1;
end
assign led = pwm_flag;

endmodule

Answer 1

第一个always语句优先级都错了。修正如下：
always@(posedge clk or negedge reset)
begin
if(!reset)
timecounter <= 0;
else if (timecounter == 26'd49999999)
timecounter <= 0;
else
timecounter <= timecounter +26'd1;
end

第二句逻辑关系有点乱，always 语句最好采用阻塞赋值，一个always语句中不要对那么多的语句赋值。Count = 16'b1000_0000_0000_0000的时候，timecounter 的值也不能刚 26'd25000000，pwm_count也不会计数。

追问

请问如果不能根据timecounter的值让pwm_count每秒加一的话那该怎么赋值呢？把count[15]==1去掉？去掉flag的赋值之后直接

begin
if(timecounter == 26'd25000000)
pwm_count = (pwm_count + 8'b00000001);
end
行么？不好意思，我是个新手，感觉做这个有点混乱。。

追答

可以的