7 classic Huawei interview questions about C language (share)

1. Find the error

void test1()
{
    char string[10];
    char* str1="0123456789";
    strcpy(string, str1);
}

The string array here is out of bounds because the string length is 10 and there is a terminator '\0' . So a total of 11 characters in length. The string array size is 10, which is out of bounds.

PS: When using the strcpy function, be sure to note that the size of the previous destination array must be larger than the size of the subsequent string, otherwise the access will be out of bounds.

void test2()
{
    char string[10], str1[10];
    for(i=0; i<10;i++)
    {
       str1[i] =&#39;a&#39;;
    }
    strcpy(string, str1);
}

There is a problem that can be seen at a glance, that is, the variable i is not defined. The compiler can help you find this during the code compilation stage, and it is easy to fix. However, many problems are self-inflicted vulnerabilities, and the compiler cannot help. The biggest problem here is that str1 has no terminator, because the second parameter of strcpy should be a string constant. This function uses the end character of the second parameter to determine whether the copy is complete. Therefore, str1p[9] = '\0' should be added after the for loop;

PS: The most obvious difference between character arrays and strings is that the string will be added with the terminator '\0' by default. '.

void test3(char* str1)
{
    char string[10];
    if(strlen(str1)<=10)
    {
       strcpy(string, str1);
    }
}

The problem here is still one of crossing the line. The strlen function gets the length of a string excluding the terminator. If it were

Summary: The above three error-finding functions mainly test the mastery of the concepts of strings and character arrays and the understanding of the strcpy function and strlen function.

2. Find errors

DSN get_SRM_no()
{
　　static int SRM_no;
　　int I;
　　for(I=0;I<MAX_SRM;I++)
    {
       SRM_no %= MAX_SRM;
       if(MY_SRM.state==IDLE)
       {
　　       break;
       }
    }
    if(I>=MAX_SRM)
       return (NULL_SRM);
    else
       return SRM_no;
}

The judgment statement of the for loop here was added by me later. It was probably lost when it was circulated on the Internet. According to The analysis of the program has been supplemented. I guess the mistake shouldn't be here.

By simply reading this function, you can roughly guess that the function of this function is to allocate a free SRAM block. Method: Detect each RAM block of SRAM starting from the RAM block after the last allocated RAM block to see if it is in the IDLE state. If it is IDLE, return the current RAM block number SRM_no. If all RAM blocks are not in the IDLE state, it means that a RAM cannot be allocated to the function caller, and a flag indicating that no RAM can be allocated (NULL_SRM) is returned.

After the above analysis, we can know that the error of this function is that the variable SRM_no is not accumulated by 1 in the for loop.

3. Write the program running result

int sum(int a)
{
    auto int c=0;
    static int b=3;
 
    c+=1;
    b+=2;
    return(a+b+c);
}
void main()
{
    int I;
    int a=2;
 
    for(I=0;I<5;I++)
    {
       printf("%d,", sum(a));
    }
}

The running result is: 8,10,12,14,16,

In the summation function c in sum is an auto variable. According to the characteristics of auto variables, variable c will be automatically assigned a value of 0 every time the sum function is called. b is a static variable. According to the characteristics of static variables, each time the sum function is called, the variable b will use the value saved by b when the sum function was last called.

A brief analysis of the function shows that if the parameters passed in do not change, the result returned each time the sum function is called is 2 more than the last time. So the answer is: 8,10,12,14,16,

4, func(1) = ?

int func(int a)
{
    int b;
    switch(a)
    {
        case 1: 30;
        case 2: 20;
        case 3: 16;
        default: 0;
    }
    return b;
}

You may have forgotten to assign a value to variable b in the case statement. If it is changed to the following code:

int func(int a)
{
    int b;
    switch(a)
    {
        case 1:      b = 30;
        case 2:      b = 20;
        case 3:      b = 16;
        default:      b = 0;
    }
    return b;
}

Because the break statement is missing from the case statement, no matter how many parameters are passed to the function, the running result is 0.

5. a[q - p] = ?

int a[3];

a[0]=0; a[1]=1; a[2]=2;

int * p, *q;

p=a;

q=&a[2];

It’s obvious: a[q - p] = a[2] = 2 ;

6. Memory space occupation problem

define int **a[3][4], then the memory space occupied by the variable is: 24 in 16-bit system, It is 48 on a 32-bit compilation system.

PS: Formula: 3 * 4 * sizeof(int **).

7. Programming

Write a function that requires the input of the year, month, day, hour, minute, and second, and outputs the next second of the year, month, day, hour, minute, and second. For example, if you enter 23:59:59 on December 31, 2004, the output will be 0:00:0 on January 1, 2005.

void ResetTheTime(int *year,int *month,int *date,int *hour,int *minute,int*second)
{
    int dayOfMonth[12]={31,28,31,30,31,30,31,31,30,31,30,31};
   
    if( *year < 0   || *month < 1 || *month > 12 ||
       *date < 1   || *date > 31 || *hour < 0   || *hour > 23 ||
       *minute < 0 ||*minute > 59|| *second <0  || *second >60 )
       return;
 
    if( *year%400 == 0 || *year%100 != 0 && *year%4 == 0 )
       dayOfMonth[1] = 29;
 
    if(*second >= 60)
    {
       *second = 0;
       *minute += 1;
       if(*minute >= 60)
       {
           *minute = 0;
           *hour += 1;
           if(*hour >= 24)
           {
              *hour = 0;
              *date += 1;
              if(*date > dayOfMonth[*month-1])
              {
                  *date = 1;
                  *month += 1;
                  if(*month > 12)
                  {
                     *month=1;
                      *year += 1;
                  }
              }
           }
       }
    }
    return;
}

Recommended: "c Language Tutorial"

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