Predict the output or
error(s) for the following:
1. void
main(){
int const * p=5;
printf("%d",++(*p));
}
Answer:
Compiler
error: Cannot modify a constant value.
Explanation:
p is a pointer to a "constant integer". But we tried to
change the value of the "constant integer".
2. main(){
char s[ ]="man";
int i;
for(i=0;s[ i ];i++)
printf("\n%c%c%c%c",s[
i ],*(s+i),*(i+s),i[s]);
}
Answer:
mmmm
aaaa
nnnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing
the same idea. Generally array name is
the base address for that array. Here s is the base address. i is
the index number/displacement from the base address. So, indirecting it with *
is same as s[i]. i[s] may be surprising. But in the case of
C it is same as s[i].
3. main(){
float me = 1.1;
double you = 1.1;
if(me==you)
printf("I love U");
else
printf("I hate
U");
}
Answer:
I hate U
Explanation:
For floating point numbers (float, double, long double) the values
cannot be predicted exactly. Depending on the number of bytes, the precision
with of the value represented varies.
Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with
less precision than long double.
Rule
of Thumb:
Never compare or at-least be cautious when using floating point
numbers with relational operators (== , >, <, <=, >=,!= ) .
4. main() {
static int var = 5;
printf("%d
",var--);
if(var)
main();
}
Answer:
5 4 3 2 1
Explanation:
When static storage class is given, it is initialized once.
The change in the value of a static variable is retained even between
the function calls. Main is also treated like any other ordinary function,
which can be called recursively.
5. main(){
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++) {
printf(" %d
",*c);
++q; }
for(j=0;j<5;j++){
printf(" %d ",*p);
++p; }
}
Answer:
2
2 2 2 2 2 3 4 6 5
Explanation:
Initially pointer c is assigned to both p and q. In
the first loop, since only q is incremented and not c , the value 2 will
be printed 5 times. In second loop p itself is incremented. So the
values 2 3 4 6 5 will be printed.
6. main(){
extern int i;
i=20;
printf("%d",i);
}
Answer:
Linker Error : Undefined symbol '_i'
Explanation:
extern storage class in the following declaration,
extern int i;
specifies to the compiler that the memory for i is
allocated in some other program and that address will be given to the current
program at the time of linking. But linker finds that no other variable of name
i is available in any other program with memory space allocated for it.
Hence a linker error has occurred .
7. main(){
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d
%d",i,j,k,l,m);
}
Answer:
0
0 1 3 1
Explanation
:
Logical operations always give a result of 1 or 0 . And
also the logical AND (&&) operator has higher priority over the logical
OR (||) operator. So the expression ‘i++
&& j++ && k++’ is executed first. The result of this
expression is 0 (-1 && -1
&& 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because
OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives
0). So the value of m is 1. The values of other variables are also incremented
by 1.
8. main(){
char *p;
printf("%d %d
",sizeof(*p),sizeof(p));
}
Answer:
1
2
Explanation:
The sizeof() operator gives the number of bytes taken by
its operand. P is a character pointer, which needs one byte for storing its
value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two
bytes to store the address of the character pointer sizeof(p) gives 2.
9. main(){
int i=3;
switch(i) {
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
}
Answer
:
three
Explanation
:
The default case can be placed anywhere inside the loop. It
is executed only when all other cases doesn't match.
10. main(){
printf("%x",-1<<4);
}
Answer:
fff0
Explanation
:
-1 is internally represented as all 1's. When left shifted four
times the least significant 4 bits are filled with 0's.The %x format specifier
specifies that the integer value be printed as a hexadecimal value.
11. main(){
char string[]="Hello
World";
display(string);
}
void
display(char *string){
printf("%s",string);
}
Answer:
Compiler Error : Type mismatch in
redeclaration of function display
Explanation :
In third line, when the function display is encountered,
the compiler doesn't know anything about the function display. It
assumes the arguments and return types to be integers, (which is the default
type). When it sees the actual function display, the arguments and type
contradicts with what it has assumed previously. Hence a compile time error
occurs.
12. main(){
int c=- -2;
printf("c=%d",c);
}
Answer:
c=2;
Explanation:
Here unary minus (or negation) operator is used twice. Same
maths rules applies, ie. minus * minus=
plus.
Note:
However you cannot give like --2. Because -- operator can only be applied to variables as a decrement
operator (eg., i--). 2 is a constant and not a variable.
13. #define int char
main(){
int i=65;
printf("sizeof(i)=%d",sizeof(i));
}
Answer:
sizeof(i)=1
Explanation:
Since the #define replaces the string int by the macro char
14. main(){
int i=10;
i=!i>14;
printf("i=%d",i);
}
Answer:
i=0
Explanation:
In the expression !i>14, NOT (!) operator has more
precedence than ‘ >’ symbol. ! is a unary logical operator. !i (!10)
is 0 (not of true is false). 0>14 is
false (zero).
15. main(){
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
77
Explanation:
p is pointing to character '\n'. str1 is pointing to character 'a'
++*p. "p is pointing to '\n' and that is incremented by one." the
ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p
is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes
'b'. ASCII value of 'b' is 98.
Now performing (11 + 98 – 32), we get 77 (77
is the ASCII value for "M");
So we get the output 77.
16. main(){
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
}
Answer:
SomeGarbageValue---1
Explanation:
p=&a[2][2][2] you
declare only two 2D arrays, but you are trying to access the third 2D(which you
are not declared) it will print garbage values. *q=***a starting address of a
is assigned integer pointer. Now q is pointing to starting address of a. If you
print *q, it will print first element of 3D array.
17. main(){
struct xx{
char name[]="hello";
};
struct xx *s;
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
You should not initialize variables in structure declaration.
18. main(){
struct xx{
int x;
struct yy{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
No output.
Explanation:
Pointer to the same type of structures are known as self
referential structures. They are particularly used in implementing
datastructures like trees. Structures within structures are known as nested
structures.
19. main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}
Answer:
hai
Explanation:
\n - newline
\b - backspace
\r - linefeed
20. main()
{
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}
Answer:
45545
Explanation:
The arguments in a function call are pushed into the stack from
left to right. The evaluation is by popping out from the stack. and the evaluation is from right to left, hence the
result.
21. #define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}
Answer:
64
Explanation:
The macro call square(4) will substituted by 4*4 so the expression
becomes i = 64/4*4 . Since / and * has equal priority the expression will be
evaluated as (64/4)*4 i.e. 16*4 = 64
22. main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s
%s",p,p1);
}
Answer:
ibj!gsjfoet
Explanation:
++*p++
will be parse in the given order
· *p
that is value at the location currently pointed by p will be taken
· ++*p
the retrieved value will be incremented
· when
; is encountered the location will be incremented, that is p++ will be executed
Hence, in the while loop initial value pointed by p is ‘h’, which
is changed to ‘i’ by executing ++*p and pointer moves to point, ‘a’ which is
similarly changed to ‘b’ and so on. Similarly blank space is converted to ‘!’.
Thus, we obtain value in p becomes “ibj!gsjfoet” and since p reaches ‘\0’ and
p1 points to p thus p1doesnot print anything.
23. #define a 10
main()
{
#define a 50
printf("%d",;
}
Answer:
50
Explanation:
The preprocessor directives can be redefined anywhere in
the program. So the most recently assigned value will be taken.
24. #define clrscr() 100
main(){
clrscr();
printf("%d\n",clrscr());
}
Answer:
100
Explanation:
Preprocessor executes as a seperate pass before the
execution of the compiler. So textual replacement of clrscr() to 100 occurs.The
input program to compiler looks like
this :
main()
{
100;
printf("%d\n",100);
}
Note:
100; is an executable statement but with no action. So it doesn't
give any problem.
25. main(){
printf("%p",main);
}
Answer:
Some address will be printed.
Explanation:
Function names are just addresses (just like array names are
addresses).main() is also a function. So the address of function main will be
printed. %p in printf specifies that the argument is an address. They are
printed as hexadecimal numbers.
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