C
Questions
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 varia 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.
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