C/C++ 综合笔试题 Part 1

1. 请写出下列代码中各变量的大小及存储位置

#define SIZE 1024
typedef struct _Node {
    struct _Node *prev;
    struct _Node *next;
    unsigned char buf[SIZE]; 
    char a;
    int ib;
    short header;
    double *p;
    unsigned char (*buf_ptr)[SIZE];
    int (*fun[3])(char (*argA)(int, int), int);
} Node;

static double m = 2.1;
int data = 0;

int main(void) {
    int *p = NULL;
    p = malloc(sizeof(int));
    char *buf = "helloworld";
    static double m = 10.0;
    Node *node = (Node*)malloc(sizeof(Node));
    char buf_stack[10];
    int i = 0;
    extern int data;
    free(p);
    free(node);
}

存储分析：

2. int i=(j=4,k=8,l=16,m=32); printf("%d", i); 输出是多少？

答案：

int i = (j = 4, k = 8, l = 16, m = 32); // 逗号表达式

逗号表达式返回最后一个表达式的值，即 m=32，因此 i=32，输出为：

32

3. 用递归方式写一个整数倒序输出函数

void print_reverse(int n) {
    if (n == 0)
        return;
    printf("%d", n % 10);
    print_reverse(n / 10);
}

示例调用：

print_reverse(1234); // 输出：4321

4. 求两个字符串的最长公共子串

char* longest_common_substring(const char* s1, const char* s2) {
    int len1 = strlen(s1);
    int len2 = strlen(s2);
    int maxlen = 0, end = 0;
    int dp[len1+1][len2+1];
    memset(dp, 0, sizeof(dp));

    for (int i = 1; i <= len1; i++) {
        for (int j = 1; j <= len2; j++) {
            if (s1[i-1] == s2[j-1]) {
                dp[i][j] = dp[i-1][j-1]+1;
                if (dp[i][j] > maxlen) {
                    maxlen = dp[i][j];
                    end = i;