C programming Interview Questions and Solution

Predict the output or error(s) for the following:

72.   main()
{
                        void swap();
            int x=10,y=8;     
                        swap(&x,&y);
            printf("x=%d y=%d",x,y);
}
void swap(int *a, int *b)
{
   *a ^= *b,  *b ^= *a, *a ^= *b; 
}          

Answer:
x=10 y=8

Explanation:
Using ^ like this is a way to swap two variables without using a temporary variable and that too in a single statement.
Inside main(), void swap(); means that swap is a function that may take any number of arguments (not no arguments) and returns nothing. So this doesn’t issue a compiler error by the call swap(&x,&y); that has two arguments. 
This convention is historically due to pre-ANSI style (referred to as Kernighan and Ritchie style) style of function declaration. In that style, the swap function will be defined as follows,
void swap()

int *a, int *b
{
   *a ^= *b,  *b ^= *a, *a ^= *b; 
}
where the arguments follow the (). So naturally the declaration for swap will look like, void swap() which means the swap can take any number of arguments.

73.      main()
{
                        int i = 257;
            int *iPtr = &i;
                        printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}

Answer:
                        1 1 

Explanation:
The integer value 257 is stored in the memory as, 00000001 00000001, so the individual bytes are taken by casting it to char * and get printed.

74.     main()
{
                        int i = 258;
            int *iPtr = &i;
                        printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}     
     
Answer:
                        2 1 

Explanation:
The integer value 257 can be represented in binary as, 00000001 00000001. Remember that the INTEL machines are ‘small-endian’ machines. Small-endian means that the lower order bytes are stored in the higher memory addresses and the higher order bytes are stored in lower addresses. The integer value 258 is stored in memory as: 00000001 00000010.  

75.     main()
{
                        int i=300;
            char *ptr = &i;
                        *++ptr=2;
            printf("%d",i);
}

Answer:
556

Explanation:
The integer value 300  in binary notation is: 00000001 00101100. It is  stored in memory (small-endian) as: 00101100 00000001. Result of the expression *++ptr = 2 makes the memory representation as: 00101100 00000010. So the integer corresponding to it  is  00000010 00101100 => 556.

76.     #include 
main()
{
char * str = "hello";
char * ptr = str;
char least = 127;
while (*ptr++)
                  least = (*ptr
printf("%d",least);
}

Answer:
0

Explanation:   
After ‘ptr’ reaches the end of the string the value pointed by ‘str’ is ‘\0’. So the value of ‘str’ is less than that of ‘least’. So the value of ‘least’ finally is 0.

77. Declare an array of N pointers to functions returning pointers to functions returning pointers to characters?
Answer:
                        (char*(*)( )) (*ptr[N])( );

78.     main()
{
struct student 
{
char name[30];
struct date dob;
}stud;
struct date
        {  
         int day,month,year;
         };
     scanf("%s%d%d%d", stud.rollno, &student.dob.day, &student.dob.month,      &student.dob.year);
}

Answer:
Compiler Error: Undefined structure date

Explanation:
Inside the struct definition of ‘student’ the member of type struct date is given. The compiler doesn’t have the definition of date structure (forward  reference is not allowed in C in this case) so it issues an error.

79.     main()
{
struct date;
struct student
{
char name[30];
struct date dob;
}stud;
struct date
            {
         int day,month,year;
 };
scanf("%s%d%d%d", stud.rollno, &student.dob.day, &student.dob.month, &student.dob.year);
}

Answer:
Compiler Error: Undefined structure date

Explanation:
Only declaration of struct date is available inside the structure definition of ‘student’ but to have a variable of type struct date the definition of the structure is required.

80.  There were 10 records stored in “somefile.dat” but the following program printed 11 names. What went wrong?
void main()
{

struct student
{          
char name[30], rollno[6];
}stud;
FILE *fp = fopen(“somefile.dat”,”r”);
while(!feof(fp))
 {
                        fread(&stud, sizeof(stud), 1 , fp);
puts(stud.name);
}
}

Explanation:
fread reads 10 records and prints the names successfully. It will return EOF only when fread tries to read another record and fails reading EOF (and returning EOF). So it prints the last record again. After this only the condition feof(fp) becomes false, hence comes out of the while loop.