---
title: MTK 題庫
description:
---
# MTK 題庫
###### tags: `Interview`
* [搭配閱讀](https://hackmd.io/yo8egq8CSYOM3S688dAaMQ?view)
## 1.
```c=
#include <stdio.h>
typedef ____(1)____;
int f1(int c) { return 1+c; }
int f2(int c) { return 2+c; }
int f3(int c) { return 3+c; }
fi g(int c){ // return to a function pointer
static fi sf[] = { f1, f2, f3 }; // declare a function pointer array
return sf[c];
}
int main(int argc, char* argv[]) {
int result = g(argc)(argc); // 假設argc=1 , g(argc)(argc) => sf[c](argc) => f2(argc)
printf("%d", result);
return 0;
}
Ans : int (*fi)(int c)
```
```c=
Re-write void (*(*papf)[3]) (char *);
typedef__________;
pf(*papf)[3];
Ans:
typedef void (*pf) (char *)
Expalin:
void (*(*papf)[3]) (char *) 原語是 declare papf as pointer to array 3 of pointer to function (pointer to char) returning void
可拆成這樣看 void (*) (char *) , (*papf)[3]
```
## 2.
```c=
#include <stdio.h>
//typedef int (*fi)(int c);
int f1(int c) { return 1+c; }
int f2(int c) { return 2+c; }
int f3(int c) { return 3+c; }
int (*g(int c))(int c){
static int (*sf[3])(int c) = { f1, f2, f3 }; // prototype = int (*f[])(int)
return sf[c];
}
int main(int argc, char* argv[]){
int result = g(argc)(argc);
printf("%d", result);
return 0;
}
```
## 3.
```c=
#include <stdio.h>
int main(int argc, char* argv[]){
enum color{red, green, blue};
typedef enum color mycolor;
mycolor m = green;
printf("%d\n",m);
return 0;
}
```
Ans: 1
## 4.
```c=
#define SUM(a,b) a+b
請問 SUM(10,1) * 5 = ?
Ans:
15
Explain:
#define SUM(a,b) (a+b) // macro只是代替字，所以有括號才是正確的
SUM(10,1) * 5 = 55
```
## 5.
```c=
#include <stdio.h>
#include <stdarg.h>
void fun(char *msg, ...);
int main()
{
fun("IndiaBIX", 1, 4, 7, 11, 0);
return 0;
}
void fun(char *msg, ...)
{
va_list ptr;
int num;
va_start(ptr, msg); // 將參數(1,4,7,11,0)放在ptr指向的一塊memory
num = va_arg(ptr, int); // va_arg取參數(以int的方式取), num = 1
num = va_arg(ptr, int); // va_arg取參數(以int的方式取), num = 4
printf("%d\n", num);
}
Ans:
4
Explain:
va_list 其實是 typedef char * va_list;
va_start 是一個macro, 把取出的指標(va_list)指到第一個不定參數，在這邊就是指到1
va_arg 是一個macro, 會將 ptr 按照 第二個參數的size移動ptr
```
## 6.
```c=
#include <stdio.h>
int f(int n)
{
return 2*n;
}
int main()
{
int n = 2;
printf("%d", f(f(f(f(f(f(f(f(n)))))))) );
return 0;
}
```
Ans: 512
## 7.
* Explain
```c=
int* (*var1) (int *(*)(int *))[10];
```
Ans : declare var1 as pointer to function (pointer to function (pointer to int) returning pointer to int) returning array 10 of pointer to int
## 8.
```c=
int cnt = 10;
const char *pc = "welcome";
while (*pc++) {
cnt++;
}
printf("%d\n", cnt);
```
Ans: 17
## 9.
```c=
請使用C語言的#define，實作
DAYS_OF_MONTH(n)
期中n可能為6~12，算出來的值必須為n月份的天數
Ans:
#define DAYS_OF_MONTH(n) \
(n == 6 || n == 9 || n == 11) ? 30 : 31
```
## 10.
Here is a c file with the code as below:
```c=
int A;
static int B;
int C[10];
void func(int D)
{
int E;
static int F;
int G[10];
...
}
For variables A,B,C,D,E,F,G:
(1) What var are used "STACK" memory?
D,E,G
(2) What's the different of B and F which are declared by "static"?
F屬於Func的全域變數
B屬於整個c的全域變數
```
## 11.
```c=
unsigned long A = 0x0001111;
unsigned long B = 0x0000202;
unsigned long C;
C = A & (~B);
C = C | B;
What's the value of C?
```
Ans 0x00001313
## 12.
```c=
#include <stdio.h>
int main()
{
int ref[] = {8, 4, 0, 2};
int *ptr;
int index;
for (index = 0, ptr = ref; index < 2; index++, ptr++) {
printf("%d %d\n", ref[index], *ptr);
}
(*ptr)++;
printf("%d %d\n", ref[index], *ptr);
return 0;
}
What's the above output value?
Ans:
8 8
4 4
1 1
```
## 13.
```c=
int *f1(void)
{
int x = 10;
return (&x); // error/warning : function returns address of local variable , 實際回傳會變成 return 0;
}
inf *f2(void)
{
int *ptr;
*ptr = 10; // pointer to address 10 鐵定是錯的
return ptr;
}
int *f3(void)
{
int *ptr;
ptr = (int *)malloc(sizeof(int));
return ptr;
}
Which of the above three functions are likely to cause problem with pointers
(a) Only f3
(b) Only f1, f3
(c) Only f1, f2
(d) f1, f2, f3
Ans: c
```
## 14.
```c=
Given an integer variable a,
a) set the bit2
b) clear the bit2
c) inverse the bit2 (0->1, 1->0)
Ans:
a. a | 4
b. a & (~4)
c. a ^ 4
```
```c=
write a code
a) set the specific bit
b) clear the specific bit
c) inverse the specific bit (0->1; 1->0)
Ans:
a. #define setBit(x, n) ((x) |= (1 << n))
b. #define clearBit(x, n) ((x) &= ~(1 << n))
c. #define inverseBit(x, n) ((x) ^= (1 << n))
```
## 15.
```c=
Consider the following program:
void e(int);
void main()
{
int a;
a = 0;
e(a);
}
void e(int n)
{
if (n < 3) {
e(++n);
printf("%d", n);
e(++n);
}
}
The output for this program is:
Ans:
3 2 1 3
```
## 16.
```c=
extern void func0(void);
extern void func1(void);
extern void func2(void);
extern void func3(void);
extern void func4(void);
extern void func5(void);
void MAIN(int N)
{
if N is 0, please execute func0(void)
if N is 1, please execute func1(void)
if N is 2, please execute func2(void)
if N is 3, please execute func3(void)
if N is 4, please execute func4(void)
if N is 5, please execute func5(void)
}
N保證有上述的6種值
請使用C實作，用你認為值型最快的方式實現
但不准用if, switch case的敘述句
Ans:
void (*f[])() = {func0, func1, func2, func3, func4, func5};
f[N]();
```
* [ref](https://medium.com/@racktar7743/c%E8%AA%9E%E8%A8%80-function-pointer%E7%9A%84%E6%87%89%E7%94%A8-%E5%9B%9B-function-pointer-array-d0d624db8406)
## 17.
```c=
extern void func0(void);
extern void func1(void);
extern void func2(void);
extern void func3(void);
extern void func4(void);
extern void func5(void);
void MAIN(int N)
{
if N is 33, please execute func0(void)
if N is 67, please execute func1(void)
if N is 194, please execute func2(void)
if N is 234, please execute func3(void)
if N is 321, please execute func4(void)
if N is 832, please execute func5(void)
}
N保證有上述的6種值
請用C的if實作，無論N為何，用你認為最快的方式實現
```
## 18.
```c=
#include <stdio.h>
int count;
int calc(int x) {
count ++;
return x > 0 ? calc(x-1) + calc(x-2) + 1 : 1; //從後面先做，也就是calc(x-2)
}
int main() {
int result = calc(5);
printf("Result %d, %d\n", result, count);
return 0;
}
Ans:
25 25
```
## 19.
```c=
請寫出下列程式之輸出結果
i = 0;
j = 10;
for (i = 0; i < 3; i++) {
j += i
}
printf("%d", j);
Ans:
13
```
## 20.
```c=
void main() {
int x = 0;
if (x = 0 || x == 0) printf("1");
if (x == 0) printf("2");
printf("3");
}
what is the output on screen?
Ans:
123
```
## 21.
```c=
What are commonly put in the stack?
(A) Local variables // stack
(B) Global variables // .bss
(C) Const data // Read Only - .data
(D) Mallloced data // heap
(E) Temporary storage for register contents // ???
(F) Jump address within a function // ???
(G) Function paremeters // stack
(h) Function return address // stack
(I) Function return value // eax
```
## 22.
```c=
What value does f() returns?
int x = 5;
int f() {
int x = 3;
{
extern int x;
return x;
}
}
Ans:
5
```
## 23.
```c=
int x;
void foo(){
int y;
}
what is the initialized value individually?
Ans:
x = 0
y = indeterminate
```
## 24.
```c=
void main()
{
char s1[] = "MTK";
char s2[] = "MTK";
if (s1 == s2)
printf("same");
else
printf("different");
}
Ans :
different
```
## 25.
```c=
Can the code be compiled?
#include <stdio.h>
#define SWAP(a, b, c)(c t; t=a, a=b, b=t)
int main()
{
int x = 10, y = 20;
SWAP(x, y, int);
printf("%d %d\n", x, y);
return 0;
}
Ans:
error: expected expression before ‘;’ token
10 | SWAP(x, y, int);
```
## 26.
```c=
#include <stdio.h>
#define INC(x) x*=2;x+=1
int main()
{
int i, j;
for (i = 0, j = 1; i < 5; ++i) // 沒括弧導致 x+=1 在macro展開的時候不被包在for裡
INC(j);
printf("Result %d", j);
return 0;
}
What's the output?
Ans:
33
```
## 27.
```c=
What is the output of the following code?
void swap(int* x, int* y)
{
int* temp=x;
x = y;
x = temp;
}
void main()
{
int a=10;
int b=20;
swap(&a,&b);
printf("%d %d", a, b);
}
Ans:
10 20
```
## 28.
```c=
#include <stdio.h>
int main(){
int goto=5;
printf("%d",goto);
return 0;
}
Ans:
Compile error
Explain:
goto是保留字
```
## 29.
```c=
void foo() {
int i = 0;
do {
++i;
if (i < 3) {
continue;
}
if (i == 5) {
break;
}
} while(0);
printf("%d\n", i);
}
Ans:
1
```
## 30.
```
What does the following declaration mean?
int (*ptr)[10];
Ans:
ptr is a pointer to an array 10 of integers
```
## 31.
```c=
Please use bitwise operation to swap two integer.
void swap(int *x, int *y) {
.......
}
Ans:
*x ^= *y;
*y ^= *x;
*x ^= *y;
```
## 32.
```c=
Given an one dimensional array of integer, please find the maxium sum of subarray.
For example:
Array = {2, 3, -8, 1, 2, 3};
Max subarray is {1, 2, 3}, sum = 6
input will consist of two lines, first line with single integer n indicates the number of elements in the array, the second line contains n space-seperated integers in the array
Ans:
int max_subarray(int *a, int n)
{
int max = a[0];
for (int i = 1; i < n; i++) {
if (max + a[i] < a[i]) {
max = a[i];
} else {
max += a[i];
}
if (r < max)
r = max;
}
return max;
}
```
## 33. Explain “#error”
```c=
寫給編譯器看的，在編譯時中斷錯誤的，例如缺少定義什麼東東。
#ifndef _CPP_
#error C++ compiler required.
#endif
```
## 34. Explain “struct” and “union”
* 在struct中，每個成員都有自己的記憶體空間
* 在union中，每個成員共用一個大空間(取最大size的成員空間)
## 35. Explain “volatile”. Can we use “const” and “volatile” in the same variable? Can we use “volatile” in a pointer?
## 36.
```c=
int a[5] ={1,2,3,4,5};
int *p = (int *)(&a+1);
ask: the value of *(a+1), (*p-1)?
Ans:
*(a+1) = 2
(*p-1) = a[5] (也就是一個unknown的stack位置)
Explain:
(int *)&a 會變成 a pointer to an array 5 of nteger
所以當 (&a+1) 的時候 p 會指向 (address of a) + (5 * sizeof(int)) 的位置
可以看成是 a[5] 的位置
```
## 37. (我不會延伸==)
* write a code that check the input is a multiple of 3 or not without using division or mod
```c=
int isMult3(unsigned int n)
{
if ( n == 0 )
return 1;
if ( n < 3 )
return 0;
n = (n >> 2) + (n & 3);
/* Recursive call till you get n
or a special terminate condition */
return(isMultN(n));
}
```
```c=
#include <stdlib.h>
int isMult3(unsigned int n) {
int odd_c = 0, even_c = 0; /* variables to count odd and even SET bits */
if (n == 0) // return true if difference is 0.
return 1;
if (n == 1) // return false if the difference is not 0.
return 0;
while (n) {
if (n & 1) // odd bit is SET, increment odd_C
odd_c++;
n >>= 1;
if (n & 1) // even bit is SET, increment even_c
even_c++;
n = n >> 1;
}
/* Recursive call till you get 0/1 */
return(isMultN(abs(odd_c - even_c)));
}
```
> 這是因為 3 可以寫成 0011 而在 3 的倍數上一直加 3 ，odd_c 、 even_c 的差值會有兩種結果出現
> 差值相等 (0011+0011 = 0110)
> 相差為 3n (10101+00011 = 11000)
## 38.
```c=
int a[5]={1,2,3,4,5};
int *p=a;
*(p++)+=123;
*(++p)+=123;
Ans:
124 2 126 4 5
Expalin
*p++ 同等於 *(p++)
```
## 39.
* Reverse a string
```c=
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define SWAP(x, y) do { \
char tmp = x; \
x = y; \
y = tmp; \
} while (0)
void revStr(char *s)
{
int len = strlen(s);
for (int i = 0; i < len/2; i++) {
SWAP(s[i], s[len-i-1]);
}
}
int main()
{
char s[] = "abcdefg";
revStr(s);
printf("%s\n", s);
}
```
## 40.
* 寫一個“標準”巨集MIN ，這個巨集輸入兩個參數並返回較小的一個。
```c=
# define MIN(x,y) ((x)<(y)?(x):(y))
```
## 41.
* 寫出一個function判斷輸入的數是2的次方
```c=
int define2N(int n) {
return (n&(n-1) == 0 && n != 0);
}
```
## 42.
* 用預處理指令#define 聲明一個常數，用以表明1年中有多少秒（忽略閏年問題）
```c=
#define SEC_PER_YEAR (60*60*24*365)
```
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