Why the output of printf("%d %d",a,*a) is 1 0 ?

Question

when *a=1 and integer in c programing..

Answer 1

make sure 'a' is defined as a pointer to another variable you've declared. like:

int b = 1;
int* a = &b;

Otherwise, if you simply declare
int* a;
*a = 1;

..then you have reserved memory for an address, but not for the contents. 'a' is pointing to a pseudo-random place in memory.

Secondly, printf() of 'a' should return the address you've reserved. printf() of '*a' should return the 1 as you're expecting. The most likely reason you're not seeing this is because 'a' is set to (1) - which means it is not a valid pointer. This can be caused by some compilers in a DEBUG build when a variable goes out of scope. As in:

int* a;
{
int b = 1;
a = &b;
}
// 'b' is now out of scope - some compilers will change memory to indicate this during a debug build. *a is now (0) !

Hope that helps.

Answer 2

it is wrong output ,
your output should be
value of 'a' and , value at 'a'
ie. (some memory address) , 1

recheck your code