1.What will be the output of the following code?
void main ()
{
int i = 0 , a[3] ;
a[i] = i++;
printf (“%d",a[i]) ;
}
Ans: The output for the above code would be a garbage value. In the statement a[i] = i++; the value of the variable i would get assigned first to a[i] i.e. a[0] and then the value of i would get incremented by 1. Since a[i] i.e. a[1] has not been initialized, a[i] will have a garbage value.
2.Why doesn't the following code give the desired result?
int x = 3000, y = 2000 ;
long int z = x * y ;
Ans: Here the multiplication is carried out between two ints x and y, and the result that would overflow would be truncated before being assigned to the variable z of type long int. However, to get the correct output, we should use an explicit cast to force long arithmetic as shown below:
long int z = ( long int ) x * y ;
Note that ( long int )( x * y ) would not give the desired effect.
more......!!!
void main ()
{
int i = 0 , a[3] ;
a[i] = i++;
printf (“%d",a[i]) ;
}
Ans: The output for the above code would be a garbage value. In the statement a[i] = i++; the value of the variable i would get assigned first to a[i] i.e. a[0] and then the value of i would get incremented by 1. Since a[i] i.e. a[1] has not been initialized, a[i] will have a garbage value.
2.Why doesn't the following code give the desired result?
int x = 3000, y = 2000 ;
long int z = x * y ;
Ans: Here the multiplication is carried out between two ints x and y, and the result that would overflow would be truncated before being assigned to the variable z of type long int. However, to get the correct output, we should use an explicit cast to force long arithmetic as shown below:
long int z = ( long int ) x * y ;
Note that ( long int )( x * y ) would not give the desired effect.
more......!!!
Monday, September 19, 2011
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