枚举类型定义了一组具有名称的值,在没有指定值时默认是int型数值。
enum {red, green, blue} light1, light2;
在上面的例子中,没有指定任何数据类型,因此使用默认的“int”数据类型。所以枚举类型light1和light2被定义为默认类型int的变量,其中包括三个成员:red, green, blue。由于没有指定各个含义的值,所以
red= 0,green= 1,blue= 2
再举一个例子
enum integer {IDLE, XX='x, S1='b01, S2='b10} state, next;
这里的数据类型是“integer”,这意味着可以将“x”和“z”值分配给枚举成员。
enum {bronze=3, silver, gold} medal;
以bronze = 3作为初始值。因此,silver等于4,gold等于5
enum {a=3, b=7, c} alphabet;
显式de为“a”和“b”赋值,因此,c将取b = 7后的值。所以c等于8,也就是说
enum {a=0, b=7, c, d=8} alphabet;
Synopsys – VCS:
Error-[ENUMDUPL] Duplicate labels in enum
The enum label 'd' has the value 4'd8 which is duplicate of enum label 'c' in the declared enum.
上面这种写法就会导致编译错误,因为c和d的值都等于8。
enum {a, b=7, c} alphabet;
“a”是第一个成员,它没有值。所以,a等于0。因此,
a = 0, b = 7, c = 8
enum bit [3:0] {red=‘h13, green, blue} color;
Synopsys – VCS:
Error-[ENUMRANGE] Enum label outside value range
The enum label 'red' has the value 'h00000013 which is outside the range of
the base type of the declared enum, which is 4 bit unsigned.
上面这个示例也会导致编译错误,因为red=‘h13超出了 bit[3:0]所能够表示的最大值('hF) 。修改成下面的写法就没有问题了。
enum bit [3:0] {red=‘d13, green, blue} color;
此时
red = 13, green = 14, and blue = 15
枚举类型在设计有限状态机时特别有用。例如,
你可以定义一个枚举类型来描述状态机的状态:
enum logic [1:0] { IDLE = 2’b00, READ = 2’b01, WRITE = 2’b10, RMW = 2’b11, ILLEGAL = ‘x } current_state,next_state;
因为类型是“logic”,所以可以将未知的('x)值分配给enum成员。这样的'x赋值对于仿真比较非常有用,而对于综合工具而言就进行合适的优化。该枚举可以在状态机编码中使用,如下所示:
always @(posedge clk, negedge reset) if (!reset) current_state <= IDLE; else current_state = next_state; always @* begin … case (current_state) IDLE : if (rdy) next_state = READ; READ : if (go) next_state = WRITE; … endcase end
Enumerated-Type Methods
枚举类型的变量提供了很多内建的方法,方便提取枚举类型成员的值。
module enum_methods; typedef enum { red, green, blue, yellow } Colors; Colors c; initial begin $display("Number of members in Colors = %0d",c.num); c = c.frst( ); $display("First member # = %0d",c); c = c.next(2); $display("c = %0d",c); c = c.last ( ); $display("Last member # = %0d",c); $display( "%s : %0d", c.name, c ); end endmodule
Simulation log:
run -all; # KERNEL: Number of members in Colors = 4 # KERNEL: First member # = 0 # KERNEL: c = 2 # KERNEL: Last member # = 3 # KERNEL: yellow : 3 # KERNEL: Simulation has fnished. There are no more test vectors to simulate. exit
在这个例子中,我们定义了一个
typedef enum {red, green, blue, yellow} Colors;
Colors有四个成员,
red= 0, green= 1, blue= 2, yellow= 3
我们首先使用“num()”方法显示枚举中的成员数量。
枚举中有四名成员。所以,显示日志显示
Number of members in Colors = 4.
然后使用方法“frst()”获取第一个成员的值。第一个成员是值为0的“red”,因此打印显示
First member = 0.
然后我们使用“next(2)”,会获取后面第2个值2.
接下来,我们通过last()获得枚举类型最后一个成员的值3
Last member # = 3
下面是另一个简单的例子:
module datatype1; enum bit [3:0] {red, green, blue=5} color; int i1; initial begin i1 = green; $display ("color.name = %s", color.name); $display ("red=%s green=%d blue=%d",color. name,green,blue); //OK $display ("red=%d green=%d blue=%d",red,green,blue); //OK end endmodule
Simulation log:
i1 = 1 color.name = red red=red green= 1 blue= 5 red= 0 green= 1 blue= 5 V C S S i m u l a t i o n R e p o r t
Enumerated Type with Ranges
枚举类型的成员可以指定范围
module datatype1; typedef enum { read=10, write[5], intr[6:8] } cycle; enum { readreg[2] = 1, writereg[2:4] = 10 } reg0; initial begin $display ("read=%0d ", read); $display ("write0=%0d write1=%0d write2=%0d write3=%0d write4=%0d ", write0,write1,write2,write3,write4); $display ("intr6=%0d, intr7=%0d intr8=%0d ",intr6, intr7, intr8); $display ("readreg0=%0d readreg1=%0d ",readreg0, readreg1); $display ("writereg2=%0d writereg3=%0d writereg4=%0d n",writereg2, writereg3, writereg4); end endmodule
typedef enum { read=10, write[5], intr[6:8] } cycle;
定义一个包含三个成员的枚举类型“cycle”。第一个“read”;然后是一系列5个writes,分别是“write0”、“write1”、“write2”、“write3”、“write4”;然后是三个“intr”的,即“intr6”、“intr7”和“intr8”。
类似的,
enum { readreg[2] = 1, writereg[2:4] = 10 } reg0;
定义一个包含两个成员:
readreg0和readreg1,
writereg2, writereg3和writereg4。
下面是仿真log
read=10 write0=11 write1=12 write2=13 write3=14 write4=15 intr6=16, intr7=17 intr8=18 readreg0=1 readreg1=2 writereg2=10 writereg3=11 writereg4=12 V C S S i m u l a t i o n R e p o r t
审核编辑:刘清
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