过程式知识表示是将有关某一问题领域的知识， 连同如何使用这些知识的方法，均隐式的表达为 一个求解问题的过程，每个过程是一段程序，完 成对具体情况的处理。过程式不像陈述式那样具有固定的形式，如何描述知识完 全取决于具体问题。
例：八数码问题 人工智能及其应用
c语言实现：
#include<stdio.h> //空格按箭头方向移动，回到起始位置 void zero_back(int start[], int array[], int *zero_place, int length) { int i; int temp; for (i = 0; i < length-1; i++) { temp = start[*zero_place]; start[*zero_place] = start[array[i + 1]]; start[array[i + 1]] = temp; *zero_place = array[i + 1]; } temp = start[*zero_place]; start[*zero_place] = start[array[0]]; start[array[0]] = temp; *zero_place = array[0]; return; } //_flag 为1则按abcdefgh方式移动；_flag不为1则使1和0不在位置上 void zero_move(int start[], int array[], int *zero_place, int length, int _flag) { int i; int j = 0; //int length; int temp; //length = sizeof(array)/sizeof(array[0]); for (i = 0; i < length; i++) { if (array[i] == *zero_place) { j = i; break; } } if (j < length - 1) { temp = start[*zero_place]; start[*zero_place] = start[array[j + 1]]; start[array[j + 1]] = temp; *zero_place = array[j + 1]; } else { if (_flag == 1) { temp = start[*zero_place]; start[*zero_place] = start[array[0]]; start[array[0]] = temp; *zero_place = array[0]; } else { temp = start[*zero_place]; start[*zero_place] = start[array[5]]; start[array[5]] = temp; *zero_place = array[5]; } } return; } void _input(int start[]) { int i; printf("请给定初始状态：\n"); printf("3*3矩阵形式输入,空格用0代替\n"); for (i = 0; i < 9; i++) { scanf("%d", &start[i]); if ((i + 1) % 3 == 0) { printf("\n"); } } } void _print(int start[]) { int i; for (i = 0; i < 9; i++) { printf("%d\t", start[i]); if ((i + 1) % 3 == 0) { printf("\n"); } } } void main() { int i; int switch_case; int start[9]; int zero_place; int array_m[9] = { 8,7,6,3,0,1,2,5,4 }; int array_a[8] = { 0,3,6,7,8,5,4,1 }; int array_b[8] = { 3,6,7,8,5,2,1,4 }; int array_c[6] = { 3,6,7,8,5,4 }; int array_d[8] = { 3,0,1,4,5,2,1,0 }; int array_e[6] = { 3,6,7,8,5,4 }; int array_f[4] = { 3,6,7,4 }; int array_g[13] = { 3,0,1,2,5,4,7,8,5,2,1,0,3 }; int array_h[4] = { 3,6,7,4 }; _input(start); for (i = 0; i < 9; i++) { if (start[i] == 0) { zero_place = i; break; } } switch_case = 1; while (switch_case < 10) { switch (switch_case) { case 1: for (i = 0; i < 50; i++) { if (start[2] != 0 && start[2] != 1) { switch_case = 2; break; } else { zero_move(start, array_m, &zero_place, 9, 0); } } printf("step 1\n"); _print(start); break; case 2: for (i = 0; i < 50; i++) { if (start[0] == 1) { switch_case = 3; break; } else { zero_move(start, array_a, &zero_place, 8, 1); } } printf("step 2\n"); _print(start); break; case 3: for (i = 0; i < 50; i++) { if (start[1] == 2) { if (start[2] == 3) { switch_case = 6; } else { switch_case = 4; } break; } else { zero_move(start, array_b, &zero_place, 8, 1); } } printf("step 3\n"); _print(start); break; case 4: for (i = 0; i < 50; i++) { if (start[4] == 3) { switch_case = 5; break; } else { zero_move(start, array_c, &zero_place, 6, 1); } } printf("step 4\n"); _print(start); break; case 5: zero_back(start, array_d, &zero_place, 8); switch_case = 6; printf("step 5\n"); _print(start); break; case 6: for (i = 0; i < 50; i++) { if (start[5] == 4) { if (start[8] == 5) { switch_case = 9; } else { switch_case = 7; } break; } else { zero_move(start, array_e, &zero_place, 6, 1); } } printf("step 6\n"); _print(start); break; case 7: for (i = 0; i < 100; i++) { if (start[4] == 5) { switch_case = 8; break; } else { zero_move(start, array_f, &zero_place, 4, 1); } } printf("step 7\n"); _print(start); break; case 8: zero_back(start, array_g, &zero_place, 13); switch_case = 9; printf("step 8\n"); _print(start); break; case 9: for (i = 0; i < 50; i++) { if (start[7] == 6&&start[4]==0) { if (start[3] == 8 && start[6] == 7) { printf("问题有解，达到目标状态\n"); } else { printf("问题无解，达不到目标状态\n"); } switch_case = 10; break; } else { zero_move(start, array_h, &zero_place, 4, 1); } } printf("step 9\n"); _print(start); break; default: break; } } } /************************** 测试情况：201 465 378 有解； **************************/ 运行结果如下： 请给定初始状态： 3*3矩阵形式输入,空格用0代替 2 0 1 4 6 5 3 7 8 step 1 2 1 5 4 6 0 3 7 8 step 2 1 6 5 0 8 7 2 4 3 step 3 1 2 8 4 0 6 3 7 5 step 4 1 2 8 0 3 4 7 5 6 step 5 1 2 3 0 4 8 7 5 6 step 6 1 2 3 7 0 4 5 6 8 step 7 1 2 3 0 5 4 6 7 8 step 8 1 2 3 0 7 4 6 8 5 问题有解，达到目标状态 step 9 1 2 3 8 0 4 7 6 5 请按任意键继续. . .