Predict the output or
error(s) for the following:
43. #ifdef something
int some=0;
#endif
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer:
Compiler error : undefined symbol some
Explanation:
This is a very simple example for conditional compilation. The name something is
not already known to the compiler making the declaration
int some = 0;
effectively removed from the source code.
44. #if something == 0
int some=0;
#endif
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer
0 0
Explanation
This code is to show that preprocessor expressions are not the same as the
ordinary expressions. If a name is not known the preprocessor treats it to be
equal to zero.
45. What is the output for the following program
main()
{
int arr2D[3][3];
printf("%d\n", ((arr2D==* arr2D)&&(* arr2D
== arr2D[0])) );
}
Answer
1
46. void main()
{
if(~0 == (unsigned int)-1)
printf(“You can answer this if you know how values are represented in memory”);
}
Answer
You can answer this if you know how values are represented in memory
Explanation
~ (tilde operator or bit-wise negation operator) operates on 0 to produce all
ones to fill the space for an integer. –1 is represented in unsigned value as
all 1’s and so both are equal.
47. int swap(int *a,int *b)
{
*a=*a+*b;*b=*a-*b;*a=*a-*b;
}
main()
{
int x=10,y=20;
swap(&x,&y);
printf("x= %d y = %d\n",x,y);
}
Answer
x = 20 y = 10
Explanation
This is one way of swapping two values. Simple checking will help understand
this.
48. main()
{
char *p = “ayqm”;
printf(“%c”,++*(p++));
}
Answer:
b
49. main()
{
int i=5;
printf("%d",++i++);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
++i yields an rvalue. For postfix ++ to operate an lvalue is required.
50. main()
{
char *p = “ayqm”;
char c;
c = ++*p++;
printf(“%c”,c);
}
Answer:
b
Explanation:
There is no difference between the expression ++*(p++) and ++*p++. Parenthesis
just works as a visual clue for the reader to see which expression is first
evaluated.
51.
int aaa() {printf(“Hi”);}
int bbb(){printf(“hello”);}
iny ccc(){printf(“bye”);}
main()
{
int ( * ptr[3]) ();
ptr[0] = aaa;
ptr[1] = bbb;
ptr[2] =ccc;
ptr[2]();
}
Answer:
bye
Explanation:
int (* ptr[3])() says that ptr is an array of pointers to functions that takes
no arguments and returns the type int. By the assignment ptr[0] = aaa; it means
that the first function pointer in the array is initialized with the address of
the function aaa. Similarly, the other two array elements also get initialized
with the addresses of
the functions bbb and ccc. Since ptr[2] contains the
address of the function ccc, the call to the function ptr[2]() is same as
calling ccc(). So it results in printing "bye".
52.
main()
{
int i=5;
printf(“%d”,i=++i ==6);
}
Answer:
1
Explanation:
The expression can be treated as i = (++i==6), because == is of higher
precedence than = operator. In the inner expression, ++i is equal to 6 yielding
true(1). Hence the result.
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