INFOSYS POWER PREPARATION
C
Questions
Note : All the programs are tested under
Turbo C/C++ compilers.
It is assumed that,
Ø Programs run under DOS environment,
Ø The underlying machine is an x86
system,
Ø 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).
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 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','\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("M");
So we get the output 77 :: "M"
(Ascii is 77).
16. #include<stdio.h>
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. #include<stdio.h>
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
You should not initialize variables in declaration
18. #include<stdio.h>
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
The structure yy is nested within structure
xx. Hence, the elements are of yy are to be
accessed through the instance of structure
xx, which needs an instance of yy to be known.
If the instance is created after defining
the structure the compiler will not know about
the instance relative to xx. Hence for nested
structure yy you have to declare member.
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.
