Why scanf must take the address of operator

Question

As the title says, I always wonder why scanf must take the address of operator (&).

Answer 1

Because C only has "pass-by-value" parameters, so to pass a 'variable' to put a value into, you have to pass its address (or a pointer to the variable).

Answer 2

scanf does not take "the address of operator (&)". It takes a pointer. Most often the pointer to the output variable is gotten by using the address-of operator in the scanf call, e.g.

int i;
scanf("%i", &i);
printf("number is: %d\n", i);

But that is not the only way to do it. The following is just as valid:

int* iPtr = malloc(sizeof(int));
scanf("%i", iPtr);
printf("number is: %d\n", *iPtr);

So is:

int i;
int iPtr = &i;
scanf("%i", iPtr);
printf("number is: %d\n", i);

Answer 3

Because it needs the address to place the value it reads. If you declare you variable as a pointer, the scanf will not need the &.

Answer 4

Everyone else has described well that sscanf needs to put its output somewhere, but why not return it? Becuase it has to return many things - it can fill in more than one variable (driven by the formatting) and it returns an int indicating how many of those variables it filled in.

Answer 5

When you input something with standard input device (usually keyboard), the data that comes through must be stored somewhere. You must point somewhere in the memory so that data can be stored there. To point a memory location, you need the address of that location. So you must pass your variable's address by using & operator with scanf().

Answer 6

as the value is going to be stored,(where?), in the memory address. so scanf() deals with (&) operator.