任务1
#include <stdio.h> #define N 4 int main() { int a[N] = {2, 0, 2, 3}; char b[N] = {'2', '0', '2', '3'}; int i; printf("sizeof(int) = %d\n", sizeof(int)); printf("sizeof(char) = %d\n", sizeof(char)); printf("\n"); // 输出int型数组a中每个元素的地址、值 for (i = 0; i < N; ++i) printf("%p: %d\n", &a[i], a[i]); printf("\n"); // 输出char型数组b中每个元素的地址、值 for (i = 0; i < N; ++i) printf("%p: %c\n", &b[i], b[i]); printf("\n"); // 输出数组名a和b对应的值 printf("a = %p\n", a); printf("b = %p\n", b); return 0; }
1.int型数组a在内存中是连续存放的,每个元素占用4个字节
2.char型数组在内存中也是连续存放的,每个元素占用1个字节
3.数组名a对应的值和&a[0]一样,数组名b对应的值和&b[0]一样
任务2
(1)
#include <stdio.h> #define N 2 #define M 3 int main() { int a[N][M] = {{1, 2, 3}, {4, 5, 6}}; char b[N][M] = {{'1', '2', '3'}, {'4', '5', '6'}}; int i, j; // 输出int型二维数组a中每个元素的地址、值 for (i = 0; i < N; ++i) for (j = 0; j < M; ++j) printf("%p: %d\n", &a[i][j], a[i][j]); printf("\n"); // 输出int型二维数组名a, 以及,a[0], a[1]的值 printf("a = %p\n", a); printf("a[0] = %p\n", a[0]); printf("a[1] = %p\n", a[1]); printf("\n"); // 输出char型二维数组b中每个元素的地址、值 for (i = 0; i < N; ++i) for (j = 0; j < M; ++j) printf("%p: %c\n", &b[i][j], b[i][j]); printf("\n"); // 输出char型二维数组名b, 以及,b[0], b[1]的值 printf("b = %p\n", b); printf("b[0] = %p\n", b[0]); printf("b[1] = %p\n", b[1]); printf("\n"); system("pause"); return 0; }
(2)
#include<stdio.h> int func(int,int); int main(){ int k=4,m=1,p1,p2; p1=func(k,m); p2=func(k,m); printf("%d,%d\n",p1,p2); return 0; } int func(int a ,int b){ static int m=0,i=2; i+=m+1; m=i+a+b; return m; }
任务3