PROGRAM: December 2008

What will print out?
main( )
{
char *p1=“name”;
char *p2;
p2=(char*)malloc(20);
memset (p2, 0, 20);
while(*p2++ = *p1++);
printf(“%s\n”,p2);
}
Answer: empty string.

What will be printed as the result of the operation below:
main()
{
int x=20,y=35;
x=y++ + x++;
y= ++y + ++x;
printf(“%d%d\n”,x,y);
}
Answer : 57 94

What will be printed as the result of the operation below:
main()
{
int x=5;
printf(“%d,%d,%d\n”,x,x< <2,x>>2);
}
Answer: 5,20,1

What will be printed as the result of the operation below:
#define swap(a,b) a=a+b;b=a-b;a=a-b;
void main()
{
int x=5, y=10;
swap (x,y);
printf(“%d %d\n”,x,y);
swap2(x,y);
printf(“%d %d\n”,x,y);
}
int swap2(int a, int b)
{
int temp;
temp=a;
b=a;
a=temp;
return 0;
}
Answer: 10, 510, 5

What will be printed as the result of the operation below:
main()
{
char *ptr = ” Cisco Systems”;
*ptr++;
printf(“%s\n”,ptr);
ptr++;
printf(“%s\n”,ptr);
}
Answer:Cisco Systems
isco systems

What will be printed as the result of the operation below:
main()
{
char s1[]=“Cisco”;
char s2[]= “systems”;
printf(“%s”,s1);
}
Answer: Cisco

What will be printed as the result of the operation below:
main()
{
char *p1;
char *p2;
p1=(char *)malloc(25);
p2=(char *)malloc(25);
strcpy(p1,”Cisco”);
strcpy(p2,“systems”);
strcat(p1,p2);
printf(“%s”,p1);
}
Answer: Ciscosystems

The following variable is available in file1.c, who can access it?
static int average;
Answer: all the functions in the file1.c can access the variable.

What will be printed as the result of the operation below:
main()
{
int x=10, y=15;
x = x++;
y = ++y;
printf(“%d %d\n”,x,y);
}
Answer: 11, 16

What will be printed as the result of the operation below:
main()
{
int a=0;
if(a==0) printf(“Cisco Systems\n”);
printf(“Cisco Systems\n”);
}
Answer: Two lines with “Cisco Systems” will be printed.

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);
}
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).

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