int *p=20;
if u print like dis printf("%d",p);
o\p:- 20; how is it possible?
plz give me the explanation.

Answers were Sorted based on User's Feedback

int *p=20; if u print like dis printf("%d",p); o\p:- 20; how is it possible? plz give m..

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.

 Is This Answer Correct ? 26 Yes 5 No

int *p=20; if u print like dis printf("%d",p); o\p:- 20; how is it possible? plz give m..

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

 Is This Answer Correct ? 8 Yes 2 No

int *p=20; if u print like dis printf("%d",p); o\p:- 20; how is it possible? plz give m..

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

 Is This Answer Correct ? 4 Yes 0 No

int *p=20; if u print like dis printf("%d",p); o\p:- 20; how is it possible? plz give m..

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;

 Is This Answer Correct ? 4 Yes 3 No

int *p=20; if u print like dis printf("%d",p); o\p:- 20; how is it possible? plz give m..

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

 Is This Answer Correct ? 1 Yes 0 No

int *p=20; if u print like dis printf("%d",p); o\p:- 20; how is it possible? plz give m..

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

 Is This Answer Correct ? 5 Yes 5 No

int *p=20; if u print like dis printf("%d",p); o\p:- 20; how is it possible? plz give m..

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

 Is This Answer Correct ? 0 Yes 0 No

int *p=20; if u print like dis printf("%d",p); o\p:- 20; how is it possible? plz give m..

/*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

 Is This Answer Correct ? 0 Yes 1 No

int *p=20; if u print like dis printf("%d",p); o\p:- 20; how is it possible? plz give m..

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.

 Is This Answer Correct ? 3 Yes 5 No

int *p=20; if u print like dis printf("%d",p); o\p:- 20; how is it possible? plz give m..

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

 Is This Answer Correct ? 1 Yes 3 No

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