Why used ampersand(&) in C/C++ for scanf method...

Question

I want to know about ampersand(&) that why used in scanf() method...

Answer 1

@Sayan Chandra, a small adjustment, an int requires 4 bytes (32 bits).

Cheers!

Answer 2

Because scanf takes a pointer as parameter, not a simple variable (you'll need to comprehend the pointer part of the course to understand it)

Answer 3

Because it's stored as pointers, you use the reference operator so you can store the result to your normal variable.

Answer 4

int a;
scanf("%d",&a);

now a is a space in your device memory

a stores whatever integer you scan from keypad / key board

as a is the memory allocated by you assiging int a(2 bytes for int)

when scanning the integer the system can't find the memory by its name that is 'a'
so u need to give its address...

address of a is &a

### as a pointer variable stores a variable's/a pointer's adress so we also say that scanf takes a pointer type as its
parameter...

### for better understanding see it as an address (which is basic)

Answer 5

ya...
i just gave him priliminary concept...
and yes
for 16 bit processor int 2 bytes
for 32/64 bit processor int 4 bytes..
👍✌lpang....

Answer 6

### also try printing the address by printf("%u",&a)
% u is for integer-adress type

remember address is always a positive integer....not 0 and not 1