Search Your Dream Job Here

Receive all updates via Facebook. Just Click the Like Button Below

C Programming Interview Questions Part Seven

C Programming Interview Questions with answers, interview questions and answers, c programming interview question part seven.learn c programming online, c programs, online c questions
 

<< Prev      1      2      3      4      5      6      7     


Part 7

100)

void main()
{
int i;
char a[]="\0";
if(printf("%s\n",a))
printf("Ok here \n");
else
printf("Forget it\n");
}

Answer:

Ok here

Explanation:

Printf will return how many characters does it print. Hence
printing a null character returns 1 which makes the if
statement true, thus "Ok here" is printed.




101)
 void main()
{
void *v;
 Page 32 6/26/2004
32
int integer=2;
int *i=&integer;
v=i;
printf("%d",(int*)*v);
}

Answer:

Compiler Error. We cannot apply indirection on type void*.

Explanation:

Void pointer is a generic pointer type. No pointer arithmetic can
be done on it. Void pointers are normally used for,
1. Passing generic pointers to functions and returning such
pointers.

2. As a intermediate pointer type.
3. Used when the exact pointer type will be known at a later
point of time.



102)
 void main()
{
int i=i++,j=j++,k=k++;
printf(“%d%d%d”,i,j,k);
}

Answer:

Garbage values.

Explanation:

An identifier is available to use in program code from the point of
its declaration.
So expressions such as i = i++ are valid statements. The i, j and k
are automatic variables and so they contain some garbage value.
Garbage in is garbage out (GIGO).




103)
 void main()
{
static int i=i++, j=j++, k=k++;
printf(“i = %d j = %d k = %d”, i, j, k);
}

Answer:

i = 1 j = 1 k = 1

Explanation:

Since static variables are initialized to zero by default.

104) void main()
{
while(1){
if(printf("%d",printf("%d")))
break;
else
continue;
}
}

Answer:

Garbage values

Explanation:

The inner printf executes first to print some garbage value. The
printf returns no of characters printed and this value also cannot be
predicted. Still the outer printf prints something and so returns a
non-zero value. So it encounters the break statement and comes
out of the while statement.



104) main()
{
unsigned int i=10;
while(i-->=0)
printf("%u ",i);
}

Answer:

10 9 8 7 6 5 4 3 2 1 0 65535 65534…..

Explanation:

Since i is an unsigned integer it can never become negative. So the
expression i-- >=0 will always be true, leading to an infinite loop.




105) #include<conio.h>
main()
{
int x,y=2,z,a;
if(x=y%2) z=2;
a=2;
printf("%d %d ",z,x);
}

Answer:

Garbage-value 0

Explanation:

The value of y%2 is 0. This value is assigned to x. The condition
reduces to if (x) or in other words if(0) and so z goes uninitialized.
Thumb Rule: Check all control paths to write bug free code.



106) main()
{
int a[10];
printf("%d",*a+1-*a+3);
}

Answer:

4

Explanation:

*a and -*a cancels out. The result is as simple as 1 + 3 = 4 !



107) #define prod(a,b) a*b
main()
{
int x=3,y=4;
printf("%d",prod(x+2,y-1));
}

Answer:

10


Explanation:

The macro expands and evaluates to as:
x+2*y-1 => x+(2*y)-1 => 10



108) main()
{
unsigned int i=65000;
while(i++!=0);
printf("%d",i);
}

Answer:

1

Explanation:

Note the semicolon after the while statement. When the value of i
becomes 0 it comes out of while loop. Due to post-increment on i
the value of i while printing is 1.




109) main()
{
int i=0;
while(+(+i--)!=0)
i-=i++;
printf("%d",i);
}

Answer:

-1

Explanation:

Unary + is the only dummy operator in C. So it has no effect on
the expression and now the while loop is, while(i--!=0) which is
false and so breaks out of while loop. The value –1 is printed due
to the post-decrement operator.




113) main()
{
float f=5,g=10;
enum{i=10,j=20,k=50};
printf("%d\n",++k);
printf("%f\n",f<<2);
printf("%lf\n",f%g);
printf("%lf\n",fmod(f,g));
}

Answer:

Line no 5: Error: Lvalue required
Line no 6: Cannot apply leftshift to float
Line no 7: Cannot apply mod to float

Explanation:

Enumeration constants cannot be modified, so you cannot apply
++.
Bit-wise operators and % operators cannot be applied on float
values.
fmod() is to find the modulus values for floats as % operator is for
ints.




110) main()
{
int i=10;
void pascal f(int,int,int);
f(i++,i++,i++);
printf(" %d",i);
}
void pascal f(integer :i,integer:j,integer :k)
{
write(i,j,k);
}

Answer:

Compiler error: unknown type integer
Compiler error: undeclared function write

Explanation:

Pascal keyword doesn’t mean that pascal code can be used. It
means that the function follows Pascal argument passing mechanism in
calling the functions.




111) void pascal f(int i,int j,int k)
{
printf(“%d %d %d”,i, j, k);
}
void cdecl f(int i,int j,int k)
{
printf(“%d %d %d”,i, j, k);
}
main()
{
int i=10;
f(i++,i++,i++);
printf(" %d\n",i);
i=10;
f(i++,i++,i++);
printf(" %d",i);
}

Answer:

10 11 12 13
12 11 10 13

Explanation:

Pascal argument passing mechanism forces the arguments to be
called from left to right. cdecl is the normal C argument passing
mechanism where the arguments are passed from right to left.




112). What is the output of the program given below
main()
{
signed char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}

Answer:

-128

Explanation:

Notice the semicolon at the end of the for loop. THe initial
value of the i is set to 0. The inner loop executes to
increment the value from 0 to 127 (the positive range of
char) and then it rotates to the negative value of -128. The
condition in the for loop fails and so comes out of the for
loop. It prints the current value of i that is -128.




113) main()
{
unsigned char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}

Answer

infinite loop

Explanation

The difference between the previous question and this one is that
the char is declared to be unsigned. So the i++ can never yield negative
value and i>=0 never becomes false so that it can come out of the for
loop.




114) main()
{
char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}

Answer:

Behavior is implementation dependent.

Explanation:

The detail if the char is signed/unsigned by default is
implementation dependent. If the implementation treats the char
to be signed by default the program will print –128 and terminate.
On the other hand if it considers char to be unsigned by default, it
goes to infinite loop.

Rule:

You can write programs that have implementation
dependent behavior. But dont write programs that depend on such
behavior.




115) Is the following statement a declaration/definition. Find what does it
mean?
int (*x)[10];

Answer

Definition.

x is a pointer to array of(size 10) integers.
Apply clock-wise rule to find the meaning of this definition.




116). What is the output for the program given below
typedef enum errorType{warning, error, exception,}error;
main()
{
error g1;
g1=1;
printf("%d",g1);
}

Answer

Compiler error: Multiple declaration for error

Explanation

The name error is used in the two meanings. One means
that it is a enumerator constant with value 1. The another use is
that it is a type name (due to typedef) for enum errorType. Given a
situation the compiler cannot distinguish the meaning of error to
know in what sense the error is used:
error g1;
g1=error;
// which error it refers in each case?
When the compiler can distinguish between usages then it
will not issue error (in pure technical terms, names can only be
overloaded in different namespaces).

Note: the extra comma in the declaration,
enum errorType{warning, error, exception,}
is not an error. An extra comma is valid and is provided just for
programmer’s convenience.




117) typedef struct error{int warning, error,
exception;}error;
main()
{
error g1;
g1.error =1;
printf("%d",g1.error);
}

Answer

1

Explanation

The three usages of name errors can be distinguishable by the
compiler at any instance, so valid (they are in different namespaces).
Typedef struct error{int warning, error, exception;}error;
This error can be used only by preceding the error by struct kayword as
in:
struct error someError;
typedef struct error{int warning, error, exception;}error;
This can be used only after . (dot) or -> (arrow) operator preceded by the
variable name as in :
g1.error =1;
printf("%d",g1.error);
typedef struct error{int warning, error, exception;}error;
This can be used to define variables without using the preceding struct
keyword as in:
error g1;
Since the compiler can perfectly distinguish between these three usages, it
is perfectly legal and valid.

Note

This code is given here to just explain the concept behind. In real
programming don’t use such overloading of names. It reduces the
readability of the code. Possible doesn’t mean that we should use it!





118) #ifdef something
int some=0;
#endif
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}

Answer:

Compiler error : undefined symbol some

Explanation:

This is a very simple example for conditional compilation.
The name something is not already known to the compiler
making the declaration
int some = 0;
effectively removed from the source code.




119) #if something == 0
int some=0;
#endif
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}

Answer

0 0

Explanation

This code is to show that preprocessor expressions are not
the same as the ordinary expressions. If a name is not
known the preprocessor treats it to be equal to zero.





120). What is the output for the following program
main()
{
int arr2D[3][3];
printf("%d\n", ((arr2D==* arr2D)&&(* arr2D == arr2D[0]))
);
}

Answer

1

Explanation

This is due to the close relation between the arrays and
pointers. N dimensional arrays are made up of (N-1)
dimensional arrays.
arr2D is made up of a 3 single arrays that contains 3
integers each .
The name arr2D refers to the beginning of all the 3 arrays.
*arr2D refers to the start of the first 1D array (of 3
integers) that is the same address as arr2D. So the
expression (arr2D == *arr2D) is true (1).
Similarly, *arr2D is nothing but *(arr2D + 0), adding a zero
doesn’t change the value/meaning. Again arr2D[0] is the
another way of telling *(arr2D + 0). So the expression
(*(arr2D + 0) == arr2D[0]) is true (1).
Since both parts of the expression evaluates to true the
result is true(1) and the same is printed.





121) void main()
{
if(~0 == (unsigned int)-1)
printf(“You can Answer this if you know how values are represented
in memory”);
}

Answer

You can answer this if you know how values are represented
in memory

Explanation

~ (tilde operator or bit-wise negation operator) operates on
0 to produce all ones to fill the space for an integer. –1 is
represented in unsigned value as all 1’s and so both are
equal.




122) int swap(int *a,int *b)
{
*a=*a+*b;*b=*a-*b;*a=*a-*b;
}
main()
{
int x=10,y=20;
swap(&x,&y);
arr2D
arr2D[1]
arr2D[2]
arr2D[3]
printf("x= %d y = %d\n",x,y);
}

Answer

x = 20 y = 10

Explanation

This is one way of swapping two values. Simple checking will help
understand this.




123) main()
{
char *p = “ayqm”;
printf(“%c”,++*(p++));
}

Answer:

b





124) main()
{
int i=5;
printf("%d",++i++);
}

Answer:

Compiler error: Lvalue required in function main

Explanation:

++i yields an rvalue. For postfix ++ to operate an lvalue is
required.




125) main()
{
char *p = “ayqm”;
char c;
c = ++*p++;
printf(“%c”,c);
}

Answer:

b

Explanation:


There is no difference between the expression ++*(p++)
and ++*p++. Parenthesis just works as a visual clue for the
reader to see which expression is first evaluated.





126)
int aaa() {printf(“Hi”);}
int bbb(){printf(“hello”);}
iny ccc(){printf(“bye”);}
main()
{
int ( * ptr[3]) ();
ptr[0] = aaa;
ptr[1] = bbb;
ptr[2] =ccc;
ptr[2]();
}

Answer:

bye

Explanation:

int (* ptr[3])() says that ptr is an array of pointers to functions
that takes no arguments and returns the type int. By the
assignment ptr[0] = aaa; it means that the first function pointer in
the array is initialized with the address of the function aaa.
Similarly, the other two array elements also get initialized with the
addresses of the functions bbb and ccc. Since ptr[2] contains the
address of the function ccc, the call to the function ptr[2]() is same
as calling ccc(). So it results in printing "bye".






127)
main()
{
int i=5;
printf(“%d”,i=++i ==6);
}

Answer:

1

Explanation:

The expression can be treated as i = (++i==6), because == is of
higher precedence than = operator. In the inner expression, ++i is
equal to 6 yielding true(1). Hence the result.





128) main()
{
char p[ ]="%d\n";
p[1] = 'c';
printf(p,65);
}

Answer:

A

Explanation:

Due to the assignment p[1] = ‘c’ the string becomes, “%c\n”.
Since this string becomes the format string for printf and ASCII
value of 65 is ‘A’, the same gets printed.





129) void ( * abc( int, void ( *def) () ) ) ();

Answer::

abc is a ptr to a function which takes 2 parameters .(a). an
integer variable.(b). a ptrto a funtion which returns void. the
return type of the function is void.

Explanation:

Apply the clock-wise rule to find the result.




130) main()
{
while (strcmp(“some”,”some\0”))
printf(“Strings are not equal\n”);
}

Answer:

No output

Explanation:

Ending the string constant with \0 explicitly makes no difference.
So “some” and “some\0” are equivalent. So, strcmp returns 0
(false) hence breaking out of the while loop.






<< Prev      PART 1      PART 2      PART 3      PART 4      PART 5      PART 6      PART 7     






Last Updated Date: April 04, 2012

0 comments:

Post a Comment

Popular Posts in this Weak