数组c与结构 - 不能改变的变量

Question

任何人都可以解释为什么

all_leds[0].pattern = 3; 

和

all_leds[1].pattern = 4; 

奈何？

#include <stdio.h> 

int main(void) 
{ 
    struct Led 
    { 
     int pin; 
     int pattern; 
    }; 

    struct Led led_1 = {1, 1}; 
    struct Led led_2 = {2, 2}; 

    printf("%d\n", led_1.pattern); // prints 1 
    printf("%d\n", led_2.pattern); // prints 2 

    struct Led all_leds[2]; 
    all_leds[0] = led_1; 
    all_leds[1] = led_2; 

    printf("%d\n", led_1.pattern); // prints 1 
    printf("%d\n", led_2.pattern); // prints 2 

    all_leds[0].pattern = 3; 
    all_leds[1].pattern = 4; 

    printf("%d\n", led_1.pattern); // prints 1 ???? 
    printf("%d\n", led_2.pattern); // prints 2 ???? 

    return 0; 
} 

Answer 1

Led是值类型（如在C的所有类型，它没有参考类型，如C++有），所以当你说all_leds[0] = led_1;你是复制在led_1结构值成all_leds第一元件。在此行之后，all_leds[0]和led_1保持独立的值，彼此之间没有任何连接。修改一个不会修改另一个。

取而代之，您可以使用指向Led值的指针填充all_leds。

struct Led * all_leds[2] = { &led_1, &led_2 }; 
// ... 
all_leds[0]->pattern = 3; 
all_leds[1]->pattern = 4; 

Answer 2

您要复制led_1和led_2值放入你的结构数组。如果你希望它们是相同的对象，你应该使你的数组​​成为指向结构体的指针数组，然后你可以通过引用来更新它们。

#include <stdio.h> 

int main(void) 
{ 
    struct Led 
    { 
     int pin; 
     int pattern; 
    }; 

    struct Led led_1 = {1, 1}; 
    struct Led led_2 = {2, 2}; 

    printf("%d\n", led_1.pattern); // prints 1 
    printf("%d\n", led_2.pattern); // prints 2 

    struct Led *all_leds[2]; 
    all_leds[0] = &led_1; 
    all_leds[1] = &led_2; 

    printf("%d\n", led_1.pattern); // prints 1 
    printf("%d\n", led_2.pattern); // prints 2 

    all_leds[0]->pattern = 3; 
    all_leds[1]->pattern = 4; 

    printf("%d\n", led_1.pattern); 
    printf("%d\n", led_2.pattern);  

    return 0; 
} 

Answer 3

好吧，我想通了这一个，它按预期工作：

#include <stdio.h> 

int main(void) 
{ 
    struct Led 
    { 
     int pin; 
     int pattern; 
    }; 

    struct Led led_1 = {1, 1}; 
    struct Led led_2 = {2, 2}; 
    printf("%d\n", led_1.pattern); 
    printf("%d\n", led_2.pattern); 

    struct Led * all_leds[2]; 
    all_leds[0] = &led_1; 
    all_leds[1] = &led_2; 

    printf("%d\n", led_1.pattern); 
    printf("%d\n", led_2.pattern); 

    all_leds[0] -> pattern = 3; 
    all_leds[1] -> pattern = 4; 
    printf("%d\n", led_1.pattern); 
    printf("%d\n", led_2.pattern); 

    return 0; 
} 

Answer 4

在这些语句

printf("%d\n", led_1.pattern); // prints 1 ???? 
printf("%d\n", led_2.pattern); // prints 2 ???? 

你ouputing结构led_1和led_2的数据成员。然而在这些陈述中

all_leds[0].pattern = 3; 
all_leds[1].pattern = 4; 

您更改了数组all_leds的元素。数组和对象led_1和Led_2是不同的对象并占用不同的内存区域。所以改变一个对象不会影响其他对象。

也许你指的是以下

printf("%d\n", all_leds[0].pattern); // prints 1 ???? 
printf("%d\n", all_leds[1].pattern); // prints 2 ???? 

或者你可以定义一个指针数组结构。例如

struct Led *all_leds[2]; 
*all_leds[0] = &led_1; 
*all_leds[1] = &led_2; 

在这种情况下，报表

all_leds[0]->pattern = 3; 
all_leds[1]->pattern = 4; 

和

printf("%d\n", led_1.pattern); // prints 1 ???? 
printf("%d\n", led_2.pattern); // prints 2 ???? 

后，你会得到预期的结果。