Note : All the programs are tested under Turbo C/C++ compilers.
It is assumed that,
· Programs run under DOS environment,
· The underlying machine is an x86 system,
· Program is compiled using Turbo C/C++ compiler.
The program output may depend on the information based on this assumptions (for example sizeof(int) == 2 may be assumed),
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;
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);
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.
131) main()
{
char str1[] = {‘s’,’o’,’m’,’e’};
char str2[] = {‘s’,’o’,’m’,’e’,’\0’};
while (strcmp(str1,str2))
printf(“Strings are not equal\n”);
}
Answer:
“Strings are not equal”
“Strings are not equal”
….
Explanation:
If a string constant is initialized explicitly with
characters, ‘\0’ is not appended automatically to the string. Since str1
doesn’t have null termination, it treats whatever the values that are in the
following positions as part of the string until it randomly reaches a ‘\0’. So
str1 and str2 are not the same, hence the result.
132) main()
{
int i = 3;
for (;i++=0;) printf(“%d”,i);
}
Answer:
Compiler Error: Lvalue
required.
Explanation:
As we know that increment operators return rvalues
and hence it cannot appear on the left
hand side of an assignment operation.
133) void main()
{
int *mptr, *cptr;
mptr = (int*)malloc(sizeof(int));
printf(“%d”,*mptr);
int *cptr = (int*)calloc(sizeof(int),1);
printf(“%d”,*cptr);
}
Answer:
garbage-value 0
Explanation:
The memory space allocated by malloc is uninitialized,
whereas calloc returns the allocated memory space initialized to zeros.
134) void main()
{
static int i;
while(i<=10)
(i>2)?i++:i--;
printf(“%d”,
i);
}
Answer:
32767
Explanation:
Since i is static it is initialized to 0. Inside the
while loop the conditional operator evaluates to false, executing i--. This
continues till the integer value rotates to positive value (32767). The while
condition becomes false and hence, comes out of the while loop, printing the i
value.
135) main()
{
int i=10,j=20;
j = i,
j?(i,j)?i:j:j;
printf("%d
%d",i,j);
}
Answer:
10 10
Explanation:
The Ternary operator ( ? : )
is equivalent for if-then-else statement. So the question can be written as:
if(i,j)
{
if(i,j)
j = i;
else
j = j;
}
else
j = j;
136) 1. const char *a;
2. char* const a;
3. char const *a;
-Differentiate the above declarations.
Answer:
1. 'const' applies to char * rather than 'a' ( pointer
to a constant char )
*a='F' : illegal
a="Hi" : legal
2. 'const' applies to 'a' rather than to the value of a (constant
pointer to char )
*a='F' : legal
a="Hi" : illegal
3. Same as 1.
137) main()
{
int i=5,j=10;
i=i&=j&&10;
printf("%d
%d",i,j);
}
Answer:
1 10
Explanation:
The expression can be written as
i=(i&=(j&&10)); The inner expression (j&&10) evaluates to 1
because j==10. i is 5. i = 5&1 is 1. Hence the result.
138) main()
{
int i=4,j=7;
j = j
|| i++ && printf("YOU CAN");
printf("%d %d", i,
j);
}
Answer:
4 1
Explanation:
The boolean expression needs to be evaluated only till
the truth value of the expression is not known. j is not equal to zero itself means that the
expression’s truth value is 1. Because it is followed by || and true ||
(anything) => true where (anything) will not be evaluated. So the
remaining expression is not evaluated and so the value of i remains the same.
Similarly when && operator is involved in an
expression, when any of the operands become false, the whole expression’s truth
value becomes false and hence the remaining expression will not be
evaluated.
false
&& (anything) => false where (anything) will not be evaluated.
139) main()
{
register int a=2;
printf("Address
of a = %d",&a);
printf("Value of a = %d",a);
}
Answer:
Compier Error: '&' on register variable
Rule to Remember:
& (address of ) operator cannot be applied
on register variables.
140) main()
{
float i=1.5;
switch(i)
{
case
1: printf("1");
case 2:
printf("2");
default :
printf("0");
}
}
Answer:
Compiler Error: switch expression not integral
Explanation:
Switch statements can be
applied only to integral types.
141) main()
{
extern i;
printf("%d\n",i);
{
int i=20;
printf("%d\n",i);
}
}
Answer:
Linker Error : Unresolved external symbol i
Explanation:
The identifier i is available in the inner block and
so using extern has no use in resolving it.
142) main()
{
int a=2,*f1,*f2;
f1=f2=&a;
*f2+=*f2+=a+=2.5;
printf("\n%d
%d %d",a,*f1,*f2);
}
Answer:
16 16 16
Explanation:
f1 and f2 both refer to the same memory location a. So
changes through f1 and f2 ultimately affects only the value of a.
143) main()
{
char *p="GOOD";
char a[
]="GOOD";
printf("\n sizeof(p) = %d, sizeof(*p) = %d,
strlen(p) = %d", sizeof(p), sizeof(*p), strlen(p));
printf("\n
sizeof(a) = %d, strlen(a) = %d", sizeof(a), strlen(a));
}
Answer:
sizeof(p) = 2, sizeof(*p) = 1, strlen(p) = 4
sizeof(a)
= 5, strlen(a) = 4
Explanation:
sizeof(p) => sizeof(char*)
=> 2
sizeof(*p)
=> sizeof(char) => 1
Similarly,
sizeof(a)
=> size of the character array => 5
When sizeof operator is applied to an array it returns
the sizeof the array and it is not
the same as the sizeof the pointer variable. Here the sizeof(a) where a is the
character array and the size of the array is 5 because the space necessary for
the terminating NULL character should also be taken into account.
144) #define DIM( array, type)
sizeof(array)/sizeof(type)
main()
{
int arr[10];
printf(“The dimension of the array is %d”, DIM(arr,
int));
}
Answer:
10
Explanation:
The size of
integer array of 10 elements is 10 * sizeof(int). The macro expands to
sizeof(arr)/sizeof(int) => 10 * sizeof(int) / sizeof(int) => 10.
145) int DIM(int array[])
{
return sizeof(array)/sizeof(int );
}
main()
{
int arr[10];
printf(“The dimension of the array is %d”,
DIM(arr));
}
Answer:
1
Explanation:
Arrays cannot be passed to functions as arguments and
only the pointers can be passed. So
the argument is equivalent to int * array (this is one of the very few places
where [] and * usage are equivalent). The return statement becomes, sizeof(int
*)/ sizeof(int) that happens to be equal in this case.
146) main()
{
static
int a[3][3]={1,2,3,4,5,6,7,8,9};
int
i,j;
static
*p[]={a,a+1,a+2};
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
printf("%d\t%d\t%d\t%d\n",*(*(p+i)+j),
*(*(j+p)+i),*(*(i+p)+j),*(*(p+j)+i));
}
}
Answer:
1 1
1 1
2 4
2 4
3 7
3 7
4 2
4 2
5 5
5 5
6 8
6 8
7 3
7 3
8 6
8 6
9 9
9 9
Explanation:
*(*(p+i)+j) is equivalent to
p[i][j].
147) 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.
148) 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.
149) 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.
150) 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.
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