真值表所有可能的组合

Question

public class truthTables {

  public static void main(String[] args) {

    boolean t = true;

    boolean f = false;

    boolean[] p = {f, f, f, f, t, t, t, t};
    boolean[] q = {f, f, t, t, f, f, f, f};
    boolean[] r = {f, t, f, t, f, t, f, t};
    boolean[] proposition = new boolean[8];
    int tableNumber = 1;

    while (tableNumber <= 256) {

      System.out.println("Table " + tableNumber + " ");
      System.out.println("P " + " q " + " r " + " proposition ");
      System.out.println("________________");

      int i = 0;

      while (i <= 7) {
        i = i;

        if (proposition[i] = true) {

          System.out.println(p[i] + "  " + q[i] + "  " + r[i] + " " + proposition[i]);
        }

        if (proposition[i] = false) {
          System.out.println(p[i] + "  " + q[i] + "  " + r[i] + " " + proposition[i]);
        }

        i = i + 1;
      }

      tableNumber = tableNumber + 1;
    }
  }
}

Answer 1

我必须创建长度为8的数组中所有可能的组合

您可以通过检查tableNumber-th表的位来创建它：

int b = 0;            // "Bit index", the element of proposition you are going to set.
int t = tableNumber;  // Copy of tableNumber that you can update.

while (t != 0) {
  proposition[b] = (t & 1) != 0;  // Is the least significant bit 1?

  b++;       // Increase the bit index.
  t >>>= 1;  // Right-shift by 1, chopping off the least significant bit.
}

您也可以将其写成for循环：

for (int b = 0, t = tableNumber; t != 0; t >>>=1, b++) {
  proposition[b] = (t & 1) != 0;
}