Mega Code Archive
INTERVIEW QUESTIONS C
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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("
%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 precession
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. #include<stdio.h>
main()
{
char s[]={'a','b','c','
','c','
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