【问题标题】:Example of a big SystemVerilog constraint大 SystemVerilog 约束的示例
【发布时间】:2012-10-30 16:09:49
【问题描述】:

你能举一个大而复杂的 SystemVerilog 约束的例子吗?越大越好,最好是现实的。也许一些地址计算也取决于其他一些变量。

我正在评估将我们的 IP 切换为使用 SystemVerilog 约束,我的管理层希望了解创建/理解 SystemVerilog 约束的难易程度。

【问题讨论】:

    标签: system-verilog


    【解决方案1】:

    由于我无法对上一篇文章发表评论,并且我建议的编辑被拒绝,我只能添加看起来像新答案但实际上并非如此的内容。叹息!

    这里是编辑 Subbdue 的示例以修复以下语法错误:

    1. SV typedef enum {...} <name>; 语法相对于 OpenVera enum <name> {...}; 语法
    2. ; 结束枚举声明
    3. 声明变量channelNumber
    4. ipVersion 声明为rand
    5. SV <var> inside { [<min>:<max>] } 语法相对于 OpenVera <var> in { min:max } 语法
    6. ; 结束constraints 中的表达式
    7. SV -> 语法相对于 OpenVera => 语法
    8. 修正错字solver -> solve
    9. 修正错字IPVx -> IPVX

    固定示例:

    class RandomConstraints;
        typedef enum {IPV4=2, IPV6, IPVX} IpVersionType;
    
        //Randomly iterate over values without repetition
        randc bit [7:0] cyclicCounter;
        //Regular random variables
        rand bit [15:0] destAddress;
        rand bit [15:0] sourceAddress;
        rand bit [15:0] numberOfPackets;
        rand bit [15:0] packetLength;
        rand bit [3:0] channelNumber;
        rand bit [7:0]  var1;
        rand bit [7:0]  var2;
        rand IpVersionType ipVersion;
    
        //Non-random variables that can be used to control constraints
        bit [15:0] minNumberOfPackets, maxNumberOfPackets;
        bit [15:0] minPacketLength, maxPacketLength;
        integer IPV4Weight, IPV6Weight;
    
        constraint cPacketLength {
            packetLength inside { [ minPacketLength : maxPacketLength ] };
        }
        constraint cChannelNumber {
            channelNumber inside {[0:1]};
        }
        // Assuming total weight adds up to 100
        constraint cIPVersionType {
            ipVersion dist { IPV4 := IPV4Weight, IPV6 := IPV6Weight,
              IPVX := (100 - IPV4Weight - IPV6Weight) };
        }
        //Probability of var1 being 1,2,3,4,5 in the ratio 1,2,4,4,4
        constraint cDist1 {
            var1 dist { 1 := 1, 2 := 2, [3:4] := 4 };
        }
        //Probability of var2 being 1,2,3,4,5 is in the ratio 1,2,4/3,4/3,4/3
        constraint cDist2 {
            var2 dist { 1 := 1, 2 := 2, [3:5] :/ 4 };
        }
        //Implication constraint - if(channelNum == i) then { ... }
        constraint cImplication {
            (channelNumber == 0) -> {
                destAddress inside {[0:200]};
                sourceAddress inside {[201:400]};
            }
            (channelNumber == 1) -> {
                destAddress inside {[201:400]};
                sourceAddress inside {[0:200]};
            }
        }
        //Controlling order of constraints solved using a constraint solver
        constraint order_solver {
            solve channelNumber before destAddress;
            solve channelNumber before sourceAddress;
        }
    
        function new();
            //Setting default min/max packets and min/max packet length
            minNumberOfPackets = 100;
            maxNumberOfPackets = 1000;
            minPacketLength = 128;
            maxPacketLength = 256;
            IPV4Weight = 50;
            IPV6Weight = 30;
        endfunction
    endclass
    

    【讨论】:

    • 干得好。看来,我需要 2k 积分才有资格审查建议的编辑(即使是在我自己的帖子上 :)).. 既然您煞费苦心地创建了一个新帖子,我将保持原样。
    【解决方案2】:

    这是一个示例,可能并不大也不复杂,但它应该让您对如何使用约束有一个现实的想法。在这里,您将看到使用的以下概念:

    1. randc 和 rand - 循环和非循环随机变量
    2. 两种类型的分布式约束
    3. 在约束中使用 if 条件,使用 隐含运算符 ... 根据另一个变量计算的地址
    4. 约束顺序求解器
    5. 在构造函数中限制随机变量的默认范围
    6. 在约束中使用枚举类型

    希望这会有所帮助......并注意在线的 cmets:

    class RandomConstraints;
        enum IpVersionType {IPV4=2, IPV6, IPVx}
    
        //Randomly iterate over values without repetition
        randc bit [7:0] cyclicCounter;
        //Regular random variables
        rand bit [15:0] destAddress;
        rand bit [15:0] sourceAddress;
        rand bit [15:0] numberOfPackets;
        rand bit [15:0] packetLength;
        rand bit [7:0]  var1;
        rand bit [7:0]  var2;
        IpVersionType ipVersion;
    
        //Non-random variables that can be used to control constraints
        bit [15:0] minNumberOfPackets, maxNumberOfPackets;
        bit [15:0] minPacketLength, maxPacketLength;
        integer IPV4Weight, IPV6Weight;
    
        constraint cPacketLength {
            packetLength in { minPacketLength : maxPacketLength }
        }
        constraint cChannelNumber {
            channelNumber in {0:1}
        }
        constraint cIPVersionType {
            ipVersion dist { IPV4 := IPV4Weight, IPV6 := IPV6Weight, IPVX := (100 - IPV4Weight - IPV6Weight) }
        }
        //Probability of var1 being 1,2,3,4,5 in the ratio 1,2,4,4,4
        constraint cDist1 {
            var1 dist { 1 := 1, 2 := 2, [3:4] := 4 }
        }
        //Probability of var2 being 1,2,3,4,5 is in the ratio 1,2,4/3,4/3,4/3 
        constraint cDist2 {
            var2 dist { 1 := 1, 2 := 2, [3:5] :/ 4 }
        }       
        //Implication constraint - if(channelNum == i) then { ... }
        constraint cImplication {
            (channelNumber == 0) => {
                destAddress in {0:200};
                sourceAddress in {201:400};
            }
            (channelNumber == 1) => {
                destAddress in {201:400};
                sourceAddress in {0:200};
            }
        }
        //Controlling order of constraints solved using a constraint solver
        constraint order_solver {
            solver channelNumber before destAddress;
            solver channelNumber before sourceAddress;
        }
    
        function new();
            //Setting default min/max packets and min/max packet length
            minNumberOfPackets = 100;
            maxNumberOfPackets = 1000;
            minPacketLength = 128;
            maxPacketLength = 256;
            IPV4Weight = 50;
            IPV6Weight = 30;
        endfunction
    endclass
    

    【讨论】:

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