C 开发学习 - 结构类型(枚举/结构/类型定义)

一、枚举

定义：枚举是 一种用户定义的数据类型，它用的关键字 enum 枚举类型名字通常并不真的使用，要用的是在大括号里地名字，因为它们就是常量符合，它们的类型是int,值则依次从0到n。 enum colors {red, yellow, green} 语法：enum 枚举类型名称{名字0m, ..., 名字n};
案例一：自动计数的枚举

//
//  main.c
//  enum
//
//  Created by liuxinming on 15/4/26.
//  Copyright (c) 2015年 liuxinming. All rights reserved.
//

#include 

enum COLOR {RED, YELLOW, GREEN, NumCOLORS};

int main(int argc, const char * argv[]) {
    int color = -1;
    char *ColorNames[NumCOLORS] = {
        red, yellow, green
    };
    char *colorName = NULL;
    
    printf(输入你喜欢的颜色代码：);
    scanf(%d, &color);
    
    if(color >= 0 && color < NumCOLORS){
        colorName = ColorNames[color];
    }else{
        colorName = unknown;
    }
    
    
    printf(你喜欢的颜色是%s
, colorName);
    return 0;
}

案例二：枚举量 声明枚举量的时候可以指定值，enum COLOR{RED=1, YELLOW, GREEN=5}

二、结构

结构是由基本数据类型构成的、并用一个标识符来命名的各种变量的组合。
结构中可以使用不同的数据类型。

申明结构的形式

struct point{
  int x;
  int y;
}

struct point p1, p2; #p1和p2都是point，里面有x和y的值

struct {
  int x;
  int y;
}p1, p2;
#p1和p2都是一种无名结构，里面有x和y的值

结构变量

struct point p; #p就是一个结构变量
p.x = 12;
p.y = 20;

案例一：使用结构体

//
//  main.c
//  structure
//
//  Created by liuxinming on 15/4/12.
//  Copyright (c) 2015年 liuxinming. All rights reserved.
//

#include 

//声明结构类型
struct date{
    int month;
    int day;
    int year;
};


int main(int argc, const char * argv[]) {
    
    //结构变量&使用
    struct date today;
    today.month = 04;
    today.day = 12;
    today.year = 2015;
    
    printf(Today is date is  %i-%i-%i.
,today.year, today.month, today.day);
    
    
    return 0;
}

案列二：结构的初始化

//
//  main.c
//  structure
//
//  Created by liuxinming on 15/4/12.
//  Copyright (c) 2015年 liuxinming. All rights reserved.
//

#include 

//声明结构类型
struct date{
    int month;
    int day;
    int year;
};


int main(int argc, const char * argv[]) {
    
    struct date today = {04, 26, 2015};
    struct date thismonty = {.month = 4, .year = 2015};
    
    printf(Today is date is  %i-%i-%i.
,today.year, today.month, today.day);
    printf(This month is  %i-%i-%i.
, thismonty.year, thismonty.month, thismonty.day);
    
    return 0;
}

输出:

Today is date is 2015-4-26.
This month is 2015-4-0.

结构成员

* 结构和数组有点像 【数组里有很多单元，结构里有很多成员 * 数组用[]运算符和下标访问其成员 a[0] = 10; * 结构用.运算符和名字访问其成员 today.day

结构运算

* 要访问整个结构，直接用结构变量的名字 * 对于整个结构，可以做赋值、取地址、也可以传递给函数参数 p1 = (struct point) {5, 10} // 相当于p1.x = 5 p1.y = 10; p1 = p2; // 相当于p1.x = p2.x ; p1.y = p2.y;

结构指针

* 和数组不同，结构变量的名字并不是结构变量的地址，必须使用&运算符 * struct date *pDate = &today;

//
//  main.c
//  structure
//
//  Created by liuxinming on 15/4/12.
//  Copyright (c) 2015年 liuxinming. All rights reserved.
//

#include 

//声明结构类型
struct date{
    int month;
    int day;
    int year;
};


int main(int argc, const char * argv[]) {
    
    struct date today;
    
    today = (struct date){04, 26, 2015};
    
    struct date day;
    
    struct date *pDate = &today;
    
    printf(Today's date is %i-%i-%i.
, today.year,today.month,today.day);
    printf(The day's date is %i-%i-%i.
, day.year, day.month, day.day);
    printf(address of today is %p
, pDate);
    
    return 0;
}

输出：

Today's date is 2015-4-26.
The day's date is 0-1606416456-32767.
address of today is 0x7fff5fbff7f0
Program ended with exit code: 0

结构作为函数参数

int numberOfDays(struct date d) * 整个结构可以作为参数的值传入函数 * 这时候是在函数内新建一个结构变量，并复制调用者的结构的值 * 也可以返回一个结构 案例：输入今天日期 ，输出明天日期【主要介绍结构体用法，不具体说明实现过程，想了解的自己研究下】

//
//  main.c
//  structure
//
//  Created by liuxinming on 15/4/12.
//  Copyright (c) 2015年 liuxinming. All rights reserved.
//

#include 
#include 

//声明结构类型
struct date{
    int month;
    int day;
    int year;
};

bool isLeap(struct date d);//判断是否为闰年
int numberOfDays(struct date d);


int main(int argc, const char * argv[]) {
    
    struct date today, tomorrow;
    
    printf(Enter today's date(mm dd yyyy):);
    scanf(%i %i %i, &today.month, &today.day, &today.year);
    
    if(today.day != numberOfDays(today)){
        tomorrow.day = today.day + 1;
        tomorrow.month = today.month;
        tomorrow.year = today.year;
    } else if (today.month == 12){
        tomorrow.day = 1;
        tomorrow.month = 1;
        tomorrow.year = today.year + 1;
    } else{
        tomorrow.day = 1;
        tomorrow.month = today.month + 1;
        tomorrow.year = today.year;
    }
    printf(Tomorrow's date is %i-%i-%i.
, tomorrow.year, tomorrow.month, tomorrow.day);
    return 0;
}

int numberOfDays(struct date d){
    int days;
    const int daysPerMonth[12] = {31,28,31,30,31,30,31,31,30,31,30,31};
    
    if(d.month == 2 && isLeap(d)){
        days = 29;//闰年
    } else{
        days = daysPerMonth[d.month - 1];
    }
    
    return days;
}

bool isLeap(struct date d){
    bool leap = false;
    if ( (d.year % 4 == 0 && d.year % 100 != 0) || d.year % 400 == 0){
        leap = true;
    }
    return leap;
}

输出：

Enter today's date(mm dd yyyy):12 31 2014
Tomorrow's date is 2015-1-1.
Program ended with exit code: 0

指向结构的指针

struct date {
  int month;
  int day;
  int year;
} myday;

struct date *p = &myday;

(*p).month = 12;
p->month = 12;

* 用->表示指针所指向的结构变量中的成员

//
//  main.c
//  structure
//
//  Created by liuxinming on 15/4/12.
//  Copyright (c) 2015年 liuxinming. All rights reserved.
//

#include 

struct point {
    int x;
    int y;
};

struct point* getStruct(struct point*);
void output(struct point);
void print(const struct point *p);


int main(int argc, const char * argv[]) {
    struct point y = {0, 0};
    getStruct(&y);
    output(y);
    output(*getStruct(&y));
    print(getStruct(&y));
    return 0;
}

struct point* getStruct(struct point *p){
    scanf(%d, &p->x);
    scanf(%d, &p->y);
    printf(%d, %d
, p->x, p->y);
    return p;
}

void output(struct point p){
    printf(%d, %d
,p.x, p.y);
}

void print(const struct point *p){
    printf(%d, %d,p->x, p->y);
}

输出：

10
50
10, 50
10, 50

结构数组

struct date dates[100];

struct date dates[] = {
 {4,5,2005},
 {2,4,2005}
};

案例：

//
//  main.c
//  structure
//
//  Created by liuxinming on 15/4/12.
//  Copyright (c) 2015年 liuxinming. All rights reserved.
//

#include 

struct time{
    int hour;
    int minutes;
    int seconds;
};

struct time timeUpdate(struct time now);


int main(void) {
   
    struct time testTime[5] = {
        {11,59,59},
        {12,0,0},
        {1,29,29},
        {23,59,59},
        {19,12,27}
    };
    int i;
    
    for (i = 0; i < 5; i++) {
        printf(Time is %.2i:%.2i:%.2i
, testTime[i].hour,testTime[i].minutes, testTime[i].seconds);
        
        testTime[i] = timeUpdate(testTime[i]);
        
        printf(... one second later it's %.2i:%.2i:%.2i
, testTime[i].hour,testTime[i].minutes, testTime[i].seconds);
    }
    
    return 0;
}

struct time timeUpdate(struct time now){
    ++ now.seconds;
    if( now.seconds == 60){
        now.seconds = 0;
        ++ now.minutes;
        
        if (now.minutes == 60){
            now.minutes = 0;
            ++ now.hour;
            
            if(now.hour == 24){
                now.hour = 0;
            }
        }
    }
    
    return now;
}

输出：

Time is 11:59:59
... one second later it's 12:00:00
Time is 12:00:00
... one second later it's 12:00:01
Time is 01:29:29
... one second later it's 01:29:30
Time is 23:59:59
... one second later it's 00:00:00
Time is 19:12:27
... one second later it's 19:12:28
Program ended with exit code: 0

结构中的结构

struct dateAndTime{
  struct date sdate;
  struct time stime;
}

嵌套的结构

struct rectangle{
    struct point pt1;
    struct point pt2;
};

    //如果有变量
    struct rectangle r;
    //就可以有：
    r.pt1.x = 1;
    r.pt1.y = 2;
    r.pt2.x = 11;
    r.pt2.y = 22;
    
    //如果有变量定义
    struct rectangle *rp;
    rp = &r;
    //那么这四种形式是等价的
    //r.pt1.x , rp->pt1.x, (r.pt1).x, (rp->pt1).x
    //但是没有rp->pt1->x 因为pt1不是指针