# struct & enum & union

## 1
What is the output of the following program?

#include‹stdio.h›
int main()
{
	struct site
	{
		char name[] = "GeeksforGeeks";
		int no_of_pages = 200;
	};
	struct site *ptr;
	printf("%d",ptr->no_of_pages);
	printf("%s",ptr->name); 
	getchar();
	return 0;
}

:::spoiler Answer
Compile Error. Note the difference between structure/union declaration and variable declaration. When you declare a structure, you actually declare a new data type suitable for your purpose. So you cannot initialize values as it is not a variable declaration but a data type declaration.

正確的寫法是：

#include <stdio.h>
#include <stdlib.h>

int main() {
  struct site {
    int no_of_pages;
  };
  struct site a = {100};
  struct site* ptr = &a;
  printf("%d", ptr->no_of_pages);
  getchar();
  return 0;
}

:::

## 2
What is the output of the following program?

int main() 
{ 
    enum{i=10,j=20,k=50}; 
    printf("%d\n",++k); 
    return 0; 
} 

:::spoiler Answer
Output:
Program will not compile and give 1 error

Explanation:
Here, i, j and k are inside the enum and therefore, they are like constants. In other words, if want to use 10 anywhere in the program, we can use i instead. In the first printf, the value of k is being modified which is not allowed because it’s enum constant.

[click here to see more information on enum](https://en.cppreference.com/w/c/language/enum)
:::

## 3

Explain “struct” and “union”

:::spoiler Answer
From [Struct declaration](https://en.cppreference.com/w/c/language/struct)

> A struct is a type consisting of a sequence of members whose storage is allocated in an ordered sequence (as opposed to union, which is a type consisting of a sequence of members whose storage overlaps).

struct 在 C 上是一種資料型態裡面可以包含多種資料型態，並且會按照程式中定義時擺放的順序在內存中排序，按照宣告的順序擺放。編譯器會自動為 struct 的成員分配空間。

在 C++ 上 struct 被擴展到可以內建 method ，跟 class 的意義非常類似，只是繼承與存取權限預設皆為 public 相反的 class 預設為 private，以及 class 有 template 的擴充 struct 沒有。

union 是一種特殊的資料型態，所有內容都從內存開頭開始存取的資料型態，並且他的大小由內部最大的那個資料型態來定義。會用同一個存儲空間，只能存儲最後一個成員訊息。只給其中一個成員給值而使用其他值就會壞掉。

補充：

enum: C語言提供自定型態為 enum，是一組由識別字所代表的整數常數（不可變更其值）。除非特別指定，不然都是由0開始，接下來遞增1。ex: 

enum week{Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,Saturday};

:::

## 4
What is the output of the following program?

#include<stdio.h> 
  
int main()
{
    struct bitfield {
        signed int a : 3;
        unsigned int b : 13;
        unsigned int c : 1;
    };
    struct bitfield bit1 = { 2, 14, 1 };
    printf("%ld\n", sizeof(bit1));
    printf("%d\n", bit1.a);
    printf("%d\n", bit1.b);
    printf("%d\n", bit1.c);
    return 0;
}

:::spoiler Answer
Output:
4
2
14
1

Explaination:

- [C 語言的 bit-field](https://hackmd.io/@sysprog/c-bitfield)

首先因為 2 對應的是 `signed int a`，所以整個 struct 會被 allocate 4 bytes of memory，但 `a` 只會站 3 個 bit，接下來 `b, c` 總共只用了 14 bits，不會超出原本既有的 4 btyes，因此 bit1 的 size 就是 4 bytes。

簡單來說，sturct 中有 bitfield 時，order matters！

- [How is the size of a struct with Bit Fields determined/measured?](https://stackoverflow.com/questions/4129961/how-is-the-size-of-a-struct-with-bit-fields-determined-measured)
:::

## 5
What is the output of the following program?

#include <stdio.h>
#define BUILD_BUG_ON_ZERO(e) (sizeof(struct { int:(-!!(e)); }))

struct foo {
    unsigned int a : 3;
    unsigned int b : 2;
    int : 0; /* Force alignment to next boundary */
    unsigned int c : 4;
    unsigned int d : 3;
};

int main() {
    int i = 0xFFFF;
    struct foo *f = (struct foo *) &i;
    printf("%d\n", sizeof(f));
    printf("a=%d\nb=%d\nc=%d\nd=%d\n", f->a, f->b, f->c, f->d);
    return 0;
}

:::spoiler Answer
Output:
8
a=7
b=3
c=4 (maybe anything)
d=6 (maybe anything)

Explaination:

- [C 語言的 bit-field](https://hackmd.io/@sysprog/c-bitfield)
:::

## 6

What is the output of the following program?

#include <stdio.h>
#define BUILD_BUG_ON_ZERO(e) (sizeof(struct { int:(-!!(e)); }))

struct foo {
    signed int a : 3;
    signed int b : 2;
    int : 0; /* Force alignment to next boundary */
    signed int c : 4;
    signed int d : 3;
};

int main() {
    int i = 0xFFFF;
    struct foo *f = (struct foo *) &i;
    printf("%d\n", sizeof(f));
    printf("a=%d\nb=%d\nc=%d\nd=%d\n", f->a, f->b, f->c, f->d);
    return 0;
}

:::spoiler Answer
Output:

8
a=-1
b=-1
c=4 (maybe anything)
d=-3 (maybe anything)

Explaination:

- [C 語言的 bit-field](https://hackmd.io/@sysprog/c-bitfield)
:::

## 7
What is the output of the following program?

#include<stdio.h> 

int main()
{
    typedef struct tag {
        char str[10];
        int a;
    } har;
 
    har h1, h2 = { "IHelp", 10 };
    h1 = h2;
    h1.str[1] = 'h';
    printf("%s, %d\n", h1.str, h1.a);
    printf("%s, %d\n", h2.str, h2.a);
    return 0;
}

:::spoiler Answer
Output:
Ihelp, 10
IHelp, 10

Explaination:

用 C++ 的 move semantics 來想就很容易了。在 move semantics 之前，都是 hard copy，而 move semantics 就像是 shallow copy。

- [Structure Assignment and Its Pitfall in C Language](http://blog.zhangliaoyuan.com/blog/2013/01/28/structure-assignment-and-its-pitfall-in-C-language/)
:::

## 8

What is the output of the following program?

#include <stdio.h>
 
struct sample {
    int a;
} sample;
 
int main()
{
    sample.a = 100;
    printf("%d", sample.a);
    return 0;
}

:::spoiler Answer
Output: 100
:::