數組c與結構 - 不能改變的變量

Question

任何人都可以解釋爲什麼

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是值類型（如在C的所有類型，它沒有參考類型，如C++有），所以當你說all_leds[0] = led_1;你是複製在led_1結構值成all_leds第一元件。在此行之後，all_leds[0]和led_1保持獨立的值，彼此之間沒有任何連接。修改一個不會修改另一個。

取而代之，您可以使用指向Led值的指針填充all_leds。

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

Answer 2

您要複製led_1和led_2值放入你的結構數組。如果你希望它們是相同的對象，你應該使你的數組​​成爲指向結構體的指針數組，然後你可以通過引用來更新它們。

#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，我想通了這一個，它按預期工作：

#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

在這些語句

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

你ouputing結構led_1和led_2的數據成員。然而在這些陳述中

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

您更改了數組all_leds的元素。數組和對象led_1和Led_2是不同的對象並佔用不同的內存區域。所以改變一個對象不會影響其他對象。

也許你指的是以下

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

或者你可以定義一個指針數組結構。例如

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

在這種情況下，報表

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 ???? 

後，你會得到預期的結果。