VHDL常量初始化

Question

我只是在研究一些VHDL代码,并且对于常量的初始化,它说如下:

constant address: integer := 16#8E#;

我以前从未见过这种定义.有人知道这是如何转换成十进制的普通数字吗？

谢谢!

Answer 1

要进一步扩展(如果您有VHDL LRM,您可以阅读第13.4.2节中的血腥细节),您可以在第#一个之前放置任何基数(在2和16之间),在第二个之后放置指数#.您也可以将_s放入单独的数字集中.

指数始终以10为基数表示,不能为负数.示例(VHDL突出显示有点不尽如人意!)

entity test_numbers is
end entity test_numbers;
architecture a1 of test_numbers is
begin 
    process is
    begin  -- process
    report integer'image(16#FF#);  -- 255
    report integer'image(10#1_000#);  -- 1000
    report integer'image(10#1#e3);  -- 1000 another way
    report integer'image(2#1#e10);  -- 1024
    report integer'image(4#100#);  -- 16
    wait;
    end process;
end architecture a1;

报告:

# Loading work.test_numbers(a1)
# ** Note: 255
#    Time: 0 ns  Iteration: 0  Instance: /test_numbers
# ** Note: 1000
#    Time: 0 ns  Iteration: 0  Instance: /test_numbers
# ** Note: 1000
#    Time: 0 ns  Iteration: 0  Instance: /test_numbers
# ** Note: 1024
#    Time: 0 ns  Iteration: 0  Instance: /test_numbers
# ** Note: 16
#    Time: 0 ns  Iteration: 0  Instance: /test_numbers