2015年11月12日 星期四

[C][Embedded System] Interview questions (面試題)

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const int a; 常整型數
int const a; 常整型數
const int *a; 整型數是不可修改的，但指標可以
int * const a; 整型數可以修改的，但指標不可以
int const * a const; 整型數不可以修改的，但指標不可以
關鍵字const的作用是為給讀你代碼的人傳達非常有用的資訊
關鍵字volatile有什麼含意?
The volatile keyword was devised to prevent compiler optimizations that might render code incorrect in the presence of certain asynchronous events.
凡牽涉讀取該volatile變數的操作，保證會到該變數的實體位址讀取

Reference:
The first should set bit 3 of a. The second should clear bit 3 of a. In both cases, the remaining bits should be unmodified.
#define BIT3 (0x1<<3)
static int a;
void set_bit3(void)
{
a |= BIT3;
}
void clear_bit3(void)
{
a &= ~BIT3;
}
Absolute address 0x67a9 to the value 0xaa55
When interview use following code is better.
int *ptr;
ptr = (int *)0x67a9;
*ptr = 0xaa55;
A more obfuscated approach is:
*(int * const)(0x67a9) = 0xaa55;
__interrupt關鍵字去定義了一個中斷服務副程式(ISR)
__interrupt double compute_area (double radius)
{
double area = PI * radius * radius;
printf(” Area = %f”, area);
return area;
}
這個函數有太多的錯誤了，以至讓人不知從何說起了：
1). ISR 不能返回一個值。
2). ISR 不能傳遞參數。
3). 在許多的處理器/編譯器中，浮點一般都是不可重入的。
#include<stdio.h>

int main()
{
    char *ptr;
    if ((ptr = (char *)malloc(0)) == NULL) 
    {
        puts("Got a null pointer");
    } else {
        puts("Got a valid pointer");
    }
}
The answer is “Got a valid pointer”
int a = 5, b = 7, c;

c = a+++b;
equal to
c = a++ + b;
explain “struct” and “union”
  • struct → 所佔記憶體空間為 member 相加
  • union → 所佔記憶體空間由最大size member決定
    note: 所以union的member同一時間只會最多出現一個 
#include <stdio.h>

int main()
{
unsigned long v1 = 0x00001111;
unsigned long v2 = 0x00001202;
unsigned long v;

v = v1 & (~v2);
printf("%x \n",v);
v = v1 | v2;
printf("%x \n",v);
}
answer
111
1313
#include <stdio.h>

int main()
{
    char str[] = "Hello";
    char *p = str;
    int n = 10;

    printf("%d %d %d",sizeof(str),sizeof(p),sizeof(n));
}
answer
6 8 4
#include <stdio.h>

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


    printf("%d %d %d %d %d",a[0],a[1],a[2],a[3],a[4]);
}
answer
129 7 131 9 10
Write down a strcmp function
int scmp(const char *pStr1,const char *pStr2)
{
    char c1, c2;
    int v;

    do {
        c1 = *pStr1++;
        c2 = *pStr2++;
        /* the casts are necessary when pStr1 is shorter & char is signed */
        v = (int)c1 - (int)c2;
    } while ((v == 0) && (c1 != '\0'));

    return v;
}

int main()
{
    scmp("happy","happy1");

    return 0;
}
List out differences between arrays and linked list
The difference between arrays and linked lists are:
  • Arrays are linear data structures. Linked lists are linear and non-linear data structures.
  • Linked lists are linear for accessing, and non-linear for storing in memory
  • Array has homogenous values. And each element is independent of each other positions. Each node in the linked list is connected with its previous node which is a pointer to the node.
  • Array elements can be modified easily by identifying the index value. It is a complex process for modifying the node in a linked list.
  • Array elements can not be added, deleted once it is declared. The nodes in the linked list can be added and deleted from the list.
Write a C code
a) Get_Bit7_of_Input
b) Set_Bit5_of_Input
c) Clear_Bit_of_Input
#include <stdio.h>

int Get_Bit7_of_Input(int in) {
    return (in & 64) >> 6;  /* Shift to first postion to show */
}
void Set_Bit5_of_Input(int *in) {
    (*in) |= 16;
}
void Clear_Bit_of_Input(int *in, int bit) {
    (*in) &= ~(1 << (bit-1));
}

int main() {
    int in1 = 65;
    printf("bit 7 of %d is %d\n", in1, Get_Bit7_of_Input(in1));

    int in2 = 0;
    int in2_set = in2;
    Set_Bit5_of_Input(&in2_set);
    printf("set bit 5 of %d to %d\n", in2, in2_set);

    int in3 = 9;
    int in3_clear = in3;
    int bit = 4;
    Clear_Bit_of_Input(&in3_clear, bit);
    printf("clear bit %d of %d to %d\n", bit, in3, in3_clear);
    return 0;
}
1. check the input is a multiple of 3

#include <stdio.h>

static int reduce(int i) {
    if (i >= 6)                /* 3 * 2 = 6 */
        return reduce((i >> 2) + (i & 0x03));
    else
        return i; // Done.
}



int a(int i){
    if (i<0) i = -i;
    i = (i >> 16) + (i & 0xFFFF);
    i = (i >> 8) + (i & 0xFF);
    i = (i >> 4) + (i & 0xF);
    i = reduce(i);
    return i == 0 || i == 3;
}

int main(){
    printf("%d",a(5));
}


2. check the input is a multiple of 5

#include <stdio.h>

static int reduce(int i) {
    if (i >= 10)                /* 3 * 2 = 6 */
        return reduce((i >> 4) + (i & 0x05));
    else
        return i; // Done.
}



int a(int i){
    if (i<0) i = -i;
    i = (i >> 16) + (i & 0xFFFF);
    i = (i >> 8) + (i & 0xFF);
    i = (i >> 4) + (i & 0xF);
    i = reduce(i);
    return i == 0 || i == 5;
}

int main(){
    printf("%d",a(195));
}

Reference :
The difference between rvalues and lvalues


An lvalue is an expression that refers to such an object.
An rvalue is any expression that has a value, but cannot have a value assigned to it


Reference:
How DMA (Direct Memory Access) work
- Direct Memory Access video Java tutorials


What is IPC
  1. Inter-process communication
  2. A detailed overview of the IPC
Linux startup process
  1. Linux startup process
Linux startup process
  1. 求職面試時，當面試官問你「有沒有問題」時，怎麼問才加分？
一行版　九九乘法表
雙變數版
 for(int i=1, j=1; i<=9; printf("%2d*%2d=%2d\n",i,j,i*j),(j==9?j=1,i++:j++));

單變數版
 for(int j=0; j<9*9; printf("%2d*%2d=%2d\n",j/9+1, j%9+1, (j/9+1)*(j%9+1)),j++);



  • [作業文]-一行版九九乘法表

    • #include <stdio.h>

int main() {
    int a = 0x12345678;
    char *ptr = (char *)&a;
    if (*ptr == 0x12)
        printf("Big Endian\n");
    else
        printf("Little Endian\n");
    return 0;
}
    Reference:
    Reference:
    C和C++的差別:
    C :具有高階語言流程控制與資料處理的便利性，以及低階語言直接(包含以位元)操作記憶體的精密性。
    C++ :C++ 兼具了效率與彈性
  • C 和 C++ 的差別 -- 一些淺見
    • Reference:

    Related Posts:

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