101) void main()
{
void *v;
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;
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;
printf("
%d\n",i);
i=10;
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);
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
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