It is assumed that,
# Programs run
under DOS environment,
# The underlying
machine is an x86 system,
# 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).
# 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.
{
int const * p=5;
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.
{
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
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.
{
float me = 1.1;
double you = 1.1;
if(me==you)
double you = 1.1;
if(me==you)
printf("I love U");
else
printf("I hate U");
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.
{
static int var = 5;
printf("%d ",var--);
if(var)
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.
{
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);
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 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);
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---10
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);
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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