ccc-test/test.cpp at main · Washington5533/ccc-test · GitHub

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#include <stdio.h>
int a[10][10] = {1};
int * init(int * p) {
	static int m = 1, n = 0;
	*p = 1 * (n / 10) * m++;
	n++;
	return p;
}
int main () {

	int m = 1;
	int c;
	scanf("%d", &c );
	switch (c ) {
		case 1: {//内存空间连续
			for (int i = 0; i < 10; i++) {
				for (int t = 0; t < 10; t++) {
					a[i][t] = i * m;
					m++;
				}
			}
			int *p;
			p = &a[0][0];
			for (int i = 0; i < 100; i++) {
				printf("%d\t", *p);
				p++;
			}
			break;
		}
		case 2: { //（函数返回指针）
			for (int i = 0; i < 10; i++) {
				for (int t = 0; t < 10; t++) {
					init(&a[i][t]);
					printf("%d\t", a[i][t]);
				}

				printf("\n");
			}
			break;

		}
		case 3 : { //（指针指向函数）
			int* (*per)(int*) = init;
			for (int i = 0; i < 10; i++) {
				for (int t = 0; t < 10; t++) {
					printf("%d\t", *(per(&a[i][t])));
				}
				printf("\n");
			}
			break;
		}
		case 4: { //二维数组的指针(a[i][j] ≡ *(a[i] + j) ≡ *(*(a + i) + j))
			for (int i = 0; i < 10; i++) {
				for (int t = 0; t < 10; t++) {
					a[i][t] = i * m;
					m++;
				}
			}
			for (int i = 0; i < 10; i++) {
				for (int t = 0; t < 10; t++) {
					printf("%d\t", *(*(a + i) + t ));
				}
				printf("\n");
			}
			break ;
		}
		case 5: {//指针数组
			for (int i = 0; i < 10; i++) {
				for (int t = 0; t < 10; t++) {
					a[i][t] = i * m;
					m++;
				}
			}
			int * pre[10];
			for (int i = 0; i < 10; i++) {
				pre[i] = a[i];
			}
			for (int i = 0; i < 10; i++) {
				for (int t = 0; t < 10; t++) {
					printf("%d\t", * (pre[i] + t));
				}
				printf("\n");
			}
			break;

		}
		case 6: {//指向一定长度数组的指针
			for (int i = 0; i < 10; i++) {
				for (int t = 0; t < 10; t++) {
					a[i][t] = i * m;
					m++;
				}
			}
			int (*pre)[10] = a;
			for (int i = 0; i < 10; i++) {
				for (int t = 0; t < 10; t ++) {
					printf("%d\t", *(pre[i] + t));
				}
				printf("\n");
			}
			break ;

		}
		case 7: { //内存读取:作为一维指针数组读取
			for (int i = 0; i < 10; i++) {
				for (int t = 0; t < 10; t++) {
					a[i][t] = i * m;
					m++;
				}
			}
			for (int i = 0; i < 10; i++) {
				for (int t = 0; t < 10; t++) {
					printf("%d\t",*(*a+i*10+t));
				}
				printf("\n");

			}
		}
	}

}