C Language Questions
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 object.
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.
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