Answer / santhi perumal

We are assigning 20 to *p. Which means we are assigning the
address 20 to p. when you want to print the address of the
pointer variable we have to print just p not *p. if you want
to print the value stored in the particular address we need
to print like *p. in this case we are printing p so it will
give the address 20 to it.

Answer / valli

int *p=20;
means
int *p;
p=20;
so the address of p is 20
printf("%d",p);
it prints 20 because now the base address of p is 20
even if we print as
printf("%u",p);
the o/p will be 20

Answer / yogi

int *p=20;
This is like int *p;p=20;
printf("%d",p);It prints p properly as 20;
printf("%d",*p);It means deference the value at address 20,
which is invalid .

If we try to run,as address 20 is invalid and it tries to
fetch the value at address 20,signal 11 sent to that process
i.e it dumps core with segmentation fault

Answer / thunder

GIVES THE ERROR DURING COMPILATION.
*P means IT CAN STORE ADDRESS NOT ANY INTEGER VALUE.

Answer / saikishore

correct ans is
int *p ; // creating a pointer of integer type
*p=20; // we are creating memory for 20 and p is pointing to
it .
printf("%d",p); // prints p 's address

printf("%d",*p); // prints value pointed by p . i.e 20

wrong declarations
we
ERRORS 1.int *p=20;

Answer / udanesh

int *p=20 means
int *p;
p=20;
so when you print the value of p definitely you will get the output as 20 because the value of p is 20

Answer / sangram

p is a pointer and it holds address.
we are assigning 20 to p;it means pointer p points the value pointed by address 20.
so to show the value on address 20 you have give *p

Answer / deepak

/*int *p=20;
is same as*/
int *p;
p=20;
so p having address of an integer value;
so
printf("%d,%u",p,p);
will give you answer 20,20

Answer / sateeshbabu aluri

we are not printing the address of variable
it mens &p;
we are printing value of p.
so,P=20 will be the o/p.

Answer / ankit

it will show compiler error as we are trying to assign integer value to pointer variable.

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