6. i 最后等于多少？

int i = 1;
int j = i++;
if ((i > j++) && (i++ == j))
    i += j;

过程分析：

• 初始：i = 1, j = i++ ⇒ j = 1, i = 2
• 判断条件：(2 > 1) ⇒ true, 再判断 (2 == 2) ⇒ true
• 所以 i += j ⇒ i = 2 + 2 = 4

最终：i = 4

7. 实现 atoi 函数

int atoi(const char *s) {
    int res = 0, sign = 1;

    while (*s == ' ') s++;
    if (*s == '-' || *s == '+') {
        if (*s == '-') sign = -1;
        s++;
    }

    while (*s >= '0' && *s <= '9') {
        res = res * 10 + (*s - '0');
        s++;
    }
    return res * sign;
}

8. 链表反转

struct Node {
    int val;
    struct Node *next;
};

struct Node* reverse(struct Node* head) {
    struct Node* prev = NULL;
    struct Node* curr = head;

    while (curr) {
        struct Node* next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next;
    }
    return prev;
}

9. 不使用库函数实现 strcat, strcmp, strcpy, strlen

int strlen(const char *s) {
    int len = 0;
    while (*s++) len++;
    return len;
}

char* strcpy(char* dst, const char* src) {
    char* ret = dst;
    while ((*dst++ = *src++));
    return ret;
}

int strcmp(const char* s1, const char* s2) {
    while (*s1 && (*s1 == *s2)) {
        s1++;
        s2++;
    }
    return *(unsigned char*)s1 - *(unsigned char*)s2;
}

char* strcat(char* dst, const char* src) {
    char* ret = dst;
    while (*dst) dst++;
    while ((*dst++ = *src++));
    return ret;
}

10. 编写 Makefile

项目结构：

.
├── main.c
├── Makefile
├── project_1
│   ├── example_1.c
│   └── example_2.c
├── project_2
    ├── example_3.c
    └── example_4.c

CC = gcc
CFLAGS = -I. -Wall

SRCS = main.c \
       project_1/example_1.c \
       project_1/example_2.c \
       project_2/example_3.c \
       project_2/example_4.c

OBJS = $(SRCS:.c=.o)

TARGET = main

$(TARGET): $(OBJS)
	$(CC) -o $@ $^

%.o: %.c
	$(CC) $(CFLAGS) -c $< -o $@

clean:
	rm -f $(OBJS) $(TARGET)

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