# 群聯 線上C測驗

## 0101(竹北)

### 1
write the following code
a.set Bit3 of int a
b.clear Bit3 of int a

a = a | (1<<3)
a = a & ~(1<<3)

---

### 2
set absolut address 0x1234 to the value 0x4321

int *p;
p = (int*) 0x1234;
*p = 0x4321;

---

### 3
下面code有何錯誤

unsigned int i=100;
for(;i>=0;i--)
{
 printf("%d\n",i);
}

* i為unsigned int，因此不會小於0，造成無窮迴圈。

#### 主管加考:
a. 如果i已經變成0，再繼續遞減會變成什麼?
* Ans : unsigned int的最大值
b. %d無法顯示到unsigned int的最大值該改成什麼?
* Ans : %lu

---

### 4
求a[]的值
int a[]={6,7,8,9,10};
int *p=a;
*(p++)+=123;
*(++p)+=123;

* a[]={129,7,131,9,10}

---

### 5
#define sum(a,b) a+b
what is sum(5,6)*10

* 5+6*10 = 65
該如何改善? 加()

---

### 6
what is size of StructA and StructB

Struct A
{
    int a;   
    short b; 
    int c;   
    char d;  
}
Struct B
{
    int a;   
    short b;  
    char c;   
    int d;   
}

* Struct A size is 16
* Struct B size is 12

---
## 0201(竹南) PCIe

### 1
* a. write a function to get bit18 value of an unsigned integer data and return 0 or 1.

`a = (x>>18) & 0x01;`

* b. write a function to set bit18 value of an unsigned integer data.

`#define get(x,n) x = x | (1 << 18)`

* c. write a function to clear bit18 value of an unsiged integer data.

`#define clear(x,n) x = x & ~(1 << 18)`

* d. write a function to inverse bit18 value of an unsiged integer data.

`#define inverse(x,n) x = x ^ (1 << 18) `

### 2 以下的迴圈印出的數字為何?
Unsigned int i;
for ( i=10 ; i>= 0; --i )
{
printf("%d\n",i);
}

* 此為無窮迴圈，因為unsigned int不會變成負值，為0時進行遞減會變成unsigned int的最大值((2^32)-1)

### 3 以linked list實作queue, 需要達到以下功能:
Push(data)：把資料放進Queue
Pop：把資料從Queue中移除
getFront：回傳最舊的資料
getBack：回傳最新的資料
IsEmpty：確認Queue裡是否有資料
getSize：回傳Queue裡的資料個數

typedef struct list
{
	int data;
	struct list *next;
}List;

List* push(List *a, int num)
{
	List *node = malloc(sizeof(struct list));
	List *first = a;
	while(a->next != NULL)
	{
		a = a ->next;
	}
	node -> data = num;
	a-> next = node;
	return first;
}

List* pop(List *a, int target)
{
	List *first = a;
	List *pre = NULL;
	while(a->data != target && a->next != NULL)
	{
		pre = a;
		a = a ->next;
	}
	a -> data = 0;
	pre->next = a->next;
	free(a);
	return first;
}

我寫了兩個也不確定對不對，主管就說往後面的題目作答了

### 4 what is the result?

int a[] = {1,2,3,4,5,6};
int *p = a;
*(p++) += 100;
*(++p) += 100;
for(inti = 0; i<6 ; i++)
printf(%d "",a[i]);" 

* a[] = {101,2,103,4,5,6}

### 5 Re-write

int(*(*abcd)[10] (char*)
typedef _______
pf(*abcd)[3]


typedef int(*pf)(char*)
這題我背考古的XD我也不知道在寫啥，但他就說好然後就下一題了。BTW，感覺最後一個應該是[10]而不是[3]

### 6 the value of *(&a+1), *(p-1)?

int a[5]={1,2,3,4,5};
int*p =(int*)(&a+1);

* *(&a+1) = 指到5之後的一個位置
* *(p-1) = 5

### 白板題1
輸入二進制的unsigned int 將其inverse，再進行輸出。

unsigned int reverse(unsigned int num)
{
	unsigned int ans = 0;
	while(num)
	{
	ans = ans<<1 + num&0x01;
	num>>1;
	}
	return ans;

}
Test case: 111100000
output : 000001111

### 白板題2
判斷符號順序合理性，如{{(<>)}}，並輸出true/false

enum apl {"{","(","<",">",")","}"}; // 0 1 2 3 4 5 此用法有誤 僅表達想法
int judge(char *s)
{	
	int len = strlen(s);
	int num = 0;
	// int lock = 0;
	int arr[100];
	int arr_pos = 0;
	for(int i=0; i< len; i++)
	{	
		if(pos == 0 && s[i]> 2)
			return 0;
		else if(s[i]<=2)
		{
			num+=s[i];
			arr[arr_pos] = s[i];
			pos++;
		}
		else if(arr[arr_pos] != 5-s[i])
			return 0;
		else if(arr[arr_pos] == 5-s[i])
		{
			num-=s[i];
			pos--;
		}		
	}
	if(num != 0)
		return 0;
	return 1;

}

正常要用stack來解，但我腦袋卡住直接忘記XD
Test case: {{(<>)}} {}()<> {{(<)>}}
output :    True     true    false

### 白板題3
不用乘除判斷是否為3的倍數，也不可以用十進制中所有數字總和為3倍數的觀念。

int mul(int num)
{	
	int odd=0,even=0;

	while(num)
	{
		odd += num&0x01;
		num>>1;
		even += num & 0x01;
		num>>1;
	}
	while(odd-even > 0)
	{
		odd -=3;
	}
	if(odd-even < 0)
		return 0;
	else if(odd-even == 0)
		return 1;
}