群聯面試

歡迎大家自行新增考古題~ 有問題的地方也可以用 "建議修訂" 標註

考古

1. 實做 C strlen, strcpy 等函式
2. 指標++ & ++指標
   https://tclin914.github.io/e9206a47/
3. 1470. Shuffle the Array
4. 給一段遞迴code print 結果
5. 20. Valid Parentheses
6. linklist insert & array reverse
7. 還有一些偏韌體ISR的問題
8. 給一段binary search code 解釋

電訪

1. n 是否為 power of 2

bool isPowerOf2(int a) {
    int tmp = a & (~a + 1);
    return tmp == a;
}

2. queue v.s. linkedlist
   https://stackoverflow.com/questions/48401777/what-is-the-difference-between-linkedlist-and-queue

群聯三題

面試

1. local, static, global
2. const, volatile
3. stack, heap 差別
   https://pjchender.dev/computer-science/cs-stack-heap/
4. function, ISR 差別
   https://www.geeksforgeeks.org/difference-between-isr-and-function-call/
5. function set bit, clear bit
6. (a & (a - 1) == 0x00) 什麼意思
   Ans :　a is power of 2
   https://stackoverflow.com/questions/600293/how-to-check-if-a-number-is-a-power-of-2
7. 依照順序填入 matrix

0	1
2	4
3	5
6	7
8	10
9	11

---

1. Assuming that we are using 32-bit compiler, pointer in 4 bytes.

int array[10] = {10, 20, 30, 40, 50, 60, 70, 80, 90, 100};
int *ptr = (int *) (&array + 1);
int *ptr2 = array;

If the address of array is 200, try to answer the following questions:
(1) &array = 200
(2) &array + 1 = 240 //跨了一整個array 所以位址+10 * sizeof(int)
(3) ptr = 240 //依照宣告定義 所以同上
(4) *ptr = 未定義 //array的下一個位址的值是多少不知道
(5) ptr - 1 = 236
(6) *(ptr - 1) = 100 //array下一個位址往前一個位址後取值 即array的最後一個值
(7) *array = 10 //array取指標 隱性轉型 指向array第一個位址的值
(8) *array + 1 = 11
(9) *(array + 1) = 20
(10) ptr2 = 200 //指向array第一個位址
(11) *ptr2 = 10 //指向array第一個位址的值
(12) ptr2[1] = 20
(13) *(ptr2 + 1) =20

2. Try to write a function that will set value “0xaa55” in the memory address of “0x1234”

int volatile * const p_reg = (int *) 0x1234; //set pointer to an address
*p_reg = 0xaa55; // dereference to set the value

https://stackoverflow.com/questions/19133962/how-to-write-a-data-in-particular-address-in-c

3. What is the output of the following code?

    void change(int *a, int &b, int c) {
            c = *a;
            b = 3;
            *a = 2;
            return;
    }
    void main() {
            int a = 1, b = 2, c = 3;
            change(&a, b, c);
            printf(“%d %d %d”, a, b, c);
            return;
    }

​​​​ **ANS: 2,3,3**

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

int get_18thbit(unsigned int n)
{
    if(n &(1<<17))
    {
        return 1;
    }
    else
    {
        return 0;
    }
}

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

int set_18thbit(unsigned int n)
{
    return n |(1<<17);
}

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

int clear_18thbit(unsigned int n)
{
    return n & ~(1<<17);
}

d. write a function to check the unsigned integer value is power of 2

bool IsPowerOfTwo(unsigned int x)
{
    return (x != 0) && ((x & (x - 1)) == 0);
}

6. What is the result?

int a[] = {1,2,3,4,5,6};
int *p = a;
*(p++) += 100; // postfix先做等號兩邊運算 再做括號內的運算 所以是p[0]+100 再位址+1到p[1]
*(++p) += 100; //prefix先做括號內的運算 再做等號兩邊運算 所以先位址+1到p[2]再取值+100
 
for(int i = 0; i<6 ; i++)
printf("%d ", a[i]);    

//補充
int *p2 = a;
int *p3 = a;
int e = (*p2)++; //e = *p = 1, 再e+1 但沒有e = e+1 所以++沒用
int f = ++*p3; //f = *p2 + 1 = 2

cout << "e =" << e << endl;
cout << "f =" << f << endl;

ANS : 101, 2, 103, 4, 5, 6

<以下兩題見Leetcode>

5. 字串反轉

補充：leetcode的func argument有string size，但先前面試經驗，需用指針實作strlen

8. linked list找中點

Other questions

Excecise 1

Exercise 2