main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}
Answer:
hai
Explanation:
\n - newline
\b - backspace
\r - linefeed
Thursday 14 March 2013
Structure error 2
#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.
Structure error example
#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
Blog Author: Vijay Kumar
Go to: Java Aptitude
Array garbage value example
#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]; // only 0 and 1 location of array can be accessed.
*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.
printf("%d",++*p + ++*str1-32);
#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");
Function error example
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.
i=!i>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).
Blog Author: Vijay Kumar
Go to: Java Aptitude
#define int char
#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
int c=- -2;
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, i.e. 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.
Blog Author: Vijay Kumar
Go to: Java Aptitude
Output in Hexadecimal
main()
{
printf("%x",-1<<4);
printf("%x",4>>-1);
}
Answer:
fff0
0fff
Explanation :
-1 is
internally represented as all 1's
1111
1111 1111 1111
When
left shifted four times the least significant 4 bits are filled with 0's.
1111
1111 1111 0000
f
f f 0
The %x
format specifier specifies that the integer value be printed as a hexadecimal
value.
switch statement with default case at top
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.
Blog Author: Vijay Kumar
Go to: Java Aptitude
sizeof(char pointer)
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 ‘p’ needs two bytes to store the address of the
character pointer sizeof(p) gives 2.
Logical Operator Example
main()
{
int
i=-1,j=-1,k=0,n=2,m;
m=i++&&j++&&k++||n++;
printf("%d
%d %d %d %d",i,j,k,n,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.
Blog Author: Vijay Kumar
Go to: Java Aptitude
extern example
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 .
Copy Array Logic
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 so first element will be printed every time, the value
2 will be printed 5 times. In second loop p
itself is incremented and printed. So the values 2 3 4 6 5 will be printed.
Calling of main() itself inside main()
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.
float==double
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 (== , >, <, <=, >=,!= ) Blog Author: Vijay Kumar
Go to: Java Aptitude
printf("\n%c%c%c%c",c[ i ],*(c+i),*(i+c),i[c]);
main()
{
char c[ ]="man";
int i;
for(i=0;c[ i ];i++)
printf("\n%c%c%c%c",c[ i
],*(c+i),*(i+c),i[c]);
}
Answer:
mmmm
aaaa
nnnn
Explanation:
c[i], *(i+c), *(c+i), i[c] are all different ways
of expressing the same idea. Generally array name is the base address for that
array. Here c 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].Blog Author: Vijay Kumar
Go to: Java Aptitude
Constant Integer example
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".
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