101.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 !
102.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.
103.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.
104.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.
105.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.
106.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.
107.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.
108.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.
109.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.
110.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.
111.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.
112.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!
113.#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.
114.#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.
115.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.
116.
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.
117.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.
118.main(){
char *p = “ayqm”;
printf(“%c”,++*(p++));
}
Answer:
b
119.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.
120.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.
121.
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".
122.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.
123.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.
124.void
( * abc( int, void ( *def) () ) ) ();
Answer:
abc is a ptr to a
function which takes 2 parameters .(. 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.
125.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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