IC学习笔记：05.SystemVerilog队列及数组方法【转载自知乎IC】

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queue_name.size                  //返回queue的大小
queue_name.insert(index,item)    //在index索引处插入item元素
queue_name.delete(index)         //刪掉某元素或整个queue
queue_name.pop_front()           //去除第一个元素(队首）
queue_name.pop_back()            //去除最后一个元素（队尾）
queue_name.push_front()          //插入元素到queue(0)（队首）
queue_name.push_back()           //插入元素到queue($)（队尾）

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int j = 1,
q2[$]={3,4}, // Queue literals do not use‘
q[$]={0,2,5}; // {0,2,5}
initial begin
  q.insert(1, j); // {0,1,2,5} Insert 1before 2
  q = {q[0:2],3,4,q[3]}; // {0,1,2,3,4,5}Insert queue in q
    q.delete(1); // {0,2,3,4,5} Delete elem.#1
// These operations are fast
  q.push_front(6);// {6,0,2,3,4,5} Insert at front
  j = q.pop_back;// {6,0,2,3,4} j = 5
  q.push_back(8);// {6,0,2,3,4,8} Insert at back
  j = q.pop_front;// {0,2,3,4,8} j = 6
  foreach (q[i])
    $display(q[i]);// Print entire queue
  q.delete(); // {} Delete entirequeue
end

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int q[$] = {2, 4, 8};    
int p[$];    
int e, pos;
e = q[0];   // 读取第一个（最左边）条目。
e = q[$];   // 读取最后一个（最右边）条目。
q[0] = e;   // 写第一个元素
p = q;   // 读和写整个队列（拷贝）
q = {q, 6};     //在队列的尾部插入'6'


q = {e, q};          // 在队列的头部插入'e'
q = q[1:$];          // 删除第一个（最左边）元素
q = q[0:$-1];          // 删除最后一个（最右边）元素
q = q[1:$-1];          // 删除第一个和最后一个元素
q = {};          // 清除队列（删除所有的元素）
q = {q[0:pos-1], e, q[pos:$]};    // 在位置'pos'处插入'e'
q = {q[0:pos], e, q[pos+1:$]};    // 在位置'pos'之后插入'e’

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 —min() //最小
 —max() //最大
 —unique() //唯一
 —find()   //returns all the elements satisfying the givenexpression.
 —find_index()  //returns the index of all the elements satisfying the given expression.
 —find_first() //returns the first element satisfying the givenexpression.
 —find_first_index()   //returns the index of the first element satisfying thegiven expression.
 —find_last()   //returns the last element satisfying the givenexpression.
 —find_last_index()    //returns the index of the last element satisfying thegiven expression.

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bit on[10]; // Array of single bits
int total;
initial begin
  foreach (on[i])
    on[i] = i;// on[i]gets 0 or 1
  // Print the single-bit sum
  $display(“on.sum =%0d”, on.sum);// on.sum =1
  // Print the sum using 32-bittotal
  $display(“on.sum =%0d”, on.sum +32’d0); // on.sum = 5
  // Sum the values using 32-bits astotal is 32-bits
  total = on.sum;
  $display("total = %0d",total); // total = 5
  // Compare the sum to a 32-bitvalue
  if (on.sum>=32’d5) // True
    $display("sum has 5 or more1’s");
    // Compute with 32-bit signed arithimetic
    $display("int sum=%0d", on.sum with (int’(item)));
end

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int count, total, d[] = ‘{9,1,8,3,4,4};
count = d.sum with (item > 7); // 2: {9, 8}
total = d.sum with ((item > 7) * item); // 17= 9+8
count = d.sum with (item < 8); // 4: {1, 3, 4, 4}
total = d.sum with (item < 8 ? item : 0); // 12=1+3+4+4
count = d.sum with (item == 4); // 2: {4, 4}

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byte b[] = ‘{ 1, 2, 3, 4};
int y;
y = b.sum ; // y becomes 10 => 1 + 2 +3 + 4
y = b.product ; // y becomes 24 => 1 * 2 *3 * 4
y = b.xor with ( item + 4 ); // y becomes12 => 5 ^ 6 ^ 7 ^ 8

logic [7:0] m [2][2] = '{ '{5, 10}, '{15, 20} };
int y;
y = m.sum with (item.sum with (item)); // y becomes 50 => 5+10+15+20

logic bit_arr[1024];
int y;
y = bit_arr.sum with ( int'(item) ); // forces result to be 32-bit
数组定位方法
int f[6] =‘{1,6,2,6,8,6};
int d[] = ‘{2,4,6,8,10};
int q[$] = {1,3,5,7}, tq[$];
tq = q.min();// {1}
tq = d.max();// {10}
tq = f.unique();// {1,6,2,8}

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int d[] = ‘{9,1,8,3,4,4}, tq[$]; // Find all elements greater than 3
tq = d.find_with (item > 3); // {9,8,4,4}
// Equivalent code
tq.delete();
foreach (d[i])
  if (d[i] > 3)
    tq.push_back(d[i]);
tq = d.find_index_with (item > 3);       //{0,2,4,5}
tq = d.find_first_with (item > 99);      // {} -nonefound
tq = d.find_first_index_with (item==8);  // {2} d[2]=8
tq = d.find_last_with (item==4);         // {4}
tq = d.find_last_index_with (item==4);   // {5} d[5]=4

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int d[] = ‘{9,1,8,3,4,4};
d.reverse();// ‘{4,4,3,8,1,9}
d.sort();// ‘{1,3,4,4,8,9}
d.rsort();// ‘{9,8,4,4,3,1}
d.shuffle();// ‘{9,4,3,8,1,4}