MTK Shenzhen company embedded software engineer pen questions (including some answers)

MTK Shenzhen Embedded Software Engineer (volume B) 1.

#define pi 3.14 #define Area(R) pi*R*R main() {  int r1=5,r2=2;  double s=0;  s=Area(r1-r2);  printf("The area is %f",s); } Result: The area is 3.700000.

2. The int compare (int a, int B) function is defined as the function pointer P of the function: int (* p) (int, int); p = compare;

3. Find the output result

#include<stdio.h> void sub(char*s,int num) {  int i ,j=num;  char t;  while(j-->1)  {    for(i=0;i<j;i++)    {      if(s[i]<s[i+1])      {        t=s[i];        s[i]=s[i+1];        s[i+1]=t;      }    }  } } main() {  char *s="CEAeded";  sub(s,6);  printf("%s\n",s) } Output result: the program crashes during running,

// 4 ************************************* *********
// Exchange the values of two variables. The third variable is not used, that is, a = 3, B = 5.
// After B = 3, a = 5
Unsigned char a = 3, B = 5;


// 5 ************************************* *************
# Define N 100
Void GetMemory1 (char * p)
{
P = (char *) malloc (sizeof (char) * N );
Strcpy (p, "Have a good day! ");
}
Char * GetMemory2 (void)
{
Char p [] = "Have a good day! ";
Return p;
}
Void main (void)
{
Char * str1 = NULL, * str2 = NULL;
GetMemory1 (str1 );
GetMemory2 (str2 );
Printf ("\ nstr1: % s", str1 );
Printf ("\ nstr2: % s", str2 );

// 6 ************************************* *****************
// Construct a single-chain table with N nodes and return the head pointer of the linked list. The sequence of each node in the linked list is required.
// Contrary to the input sequence of node data, for example, input 1, 2, 3, 4, and 5, the linked list is
// Head-> 5 4 3 2 1, supplemental procedure
# Define N 10
Typedef struct Node
{
Int data;
Struct Node * next;
} NODE;
Int Get_Data (int I); // The definition is omitted.
Node * Create_u ()
{
Int I;
NODE * p, * Head = NULL;
For (I = 0; I <N; I ++)
{
VP = New NODE;
P-> Data = Get_Data (I );
________________;
________________;
}
Return Head;
}

It's actually a head plug-in.

// 7 ************************************* *********
// N node linked lists. Each node stores one character to determine whether the characters stored in the linked list are
// Symmetric center, that is, a B c B a or a B c B a. Supplementary procedure
Typedef struct Node
{
Int data;
Struct Node * next;
} NODE;
Bool Is_symmeic (NODE * head, * int n)
{
Char D [N];
Int I, d;
__________;
For (I = 0; I <d; I ++)
{
D [I] = head-> data;
Head = head-> next;
}
If (__________)
{
Head = head-> next;
}
While (head)
{
_______________;
If (D [I]! = Head-> data)
{
Return false;
}
Head = head-> next;
}
Return true;
}
// 8 *************************************
// The str function contains only uppercase and lowercase letters. The change_move (char * str) function is used to convert uppercase letters to uppercase letters * and
// Move to the front case and return the number *
// If AabBdcYY is changed to ***** abd, 5 is returned.
Int chang_move (char * str)
{
Int len, I, curstr =-1;
Len = strlen (str );
For (I = len-1; I> = 0; I --)
{
If (str [I]> = 'A' & str [I] <= 'Z ')
{
Str [I] = '*';
If (cursor =-1)
{
Cursor = I;
}
Else if (cursor> I)
{
_____________;
Str [I] = '*';
_____________;

}
}
Return ____________;
}
// 9 ************************************* **********
// Obtain the first public substring of two strings and return the substring
// For example, "a B c d e f g e h I" "a c d e f ** g e h I"
// The first one is "c d e f"; strcmp () is not allowed ()
Char * Maxf (char * str1, char * str2)
{
}