Simple C Language Question and answer

1)

main()

{

int a[10];

printf("%d",*a+1-*a+3);

}

Answer:

4  

Explanation:

*a and -*a cancels out. The result is as simple as 1 + 3 = 4 !

2)

#define prod(a,b) a*b

main() 

{

int x=3,y=4;

printf("%d",prod(x+2,y-1));

}

Answer:-

10

Explanation:

The macro expands and evaluates to as:

x+2*y-1 => x+(2*y)-1 => 10

3)

main()

{

unsigned int i=65000;

while(i++!=0);

printf("%d",i);

}

Answer:-

 1

Explanation:

Note the semicolon after the while statement. When the value of i becomes 0 it comes out of while loop. Due to post-increment on i the value of i while printing is 1.

  

4)

main()

{

int i=0;

while(+(+i--)!=0)

i- =i++;

printf("%d",i);

}

Answer:-

-1

Explanation:

Unary + is the only dummy operator in C. So it has no effect on the expression and now the while loop is, while(i--!=0) which is false and so breaks out of while loop. The value –1 is printed due to the post-decrement operator.

 

5)

main()

{

float f=5,g=10;

enum{i=10,j=20,k=50};

printf("%d\n",++k);

printf("%f\n",f<<2);

printf("%lf\n",f%g);

printf("%lf\n",fmod(f,g)); 

}

Answer:

Line no 5: Error: Lvalue required

Line no 6: Cannot apply leftshift to float

Line no 7: Cannot apply mod to float

Explanation:

Enumeration constants cannot be modified, so you cannot apply ++.

Bit-wise operators and % operators cannot be applied on float values.

fmod() is to find the modulus values for floats as % operator is for ints.  

6)  

main()

{

int i=10;

void pascal f(int,int,int);

f(i++,i++,i++);

printf(" %d",i);

}

void pascal f(integer :i,integer:j,integer :k)

{ 

write(i,j,k); 

}

Answer:-

Compiler error:  unknown type integer

Compiler error:  undeclared function write

Explanation:

Pascal keyword doesn’t mean that pascal code can be used. It means that the function follows Pascal argument passing mechanism in calling the functions. 

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