Thread: Pointers: & and * when scanning

Hello, I have a quick question about using & and * when it comes to pointers and scanf.

If I have

Code:

int *ptr;
int a = 3, b;

What happens if I have

Code:

ptr = &b;
scanf("%d", &ptr);

where there's a & in front of the pointer

or

Code:

ptr = &b;
scanf("%d", *ptr);

where there's a * in front of the pointer?

I know that when you scan something into an array, you don't need the & because it's a pointer. So that means that when you scan something into a pointer, you don't need it. But what happens if you have a & or a * in front of ptr? Does * mean that the value that is scanned in s put into b?

Thank so much for your answers!

So in the case of scanf, which is what I have up there, where am I scanning in values when I use &ptr or *ptr.
I'm not printing in this case.

It's actually undefined what will happen, because your object types, *ptr and &ptr, do not match your conversion specifier, %d.

But to answer your question, suppose scanf() does actually attempt to store the converted input, and you entered 55:

With the correct statement,

Code:

scanf("%d", ptr);

the value 55 would be stored in b.

With &ptr as an argument, it will try to store the value 55 in ptr.
Now ptr, in all likelyhood, no longer points to b.

With *ptr as an argument, it will try to store the value 55 in whatever b is pointing to.
This is where the real havoc can be wrought.

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Originally Posted by GReaper

"&ptr" gives the leftmost value
"ptr" gives the rightmost value
"*ptr" gives the rightmost value of the address that exists at its own rightmost value( it points to the value ). You can visualize it as:
0x100000 -> 0x100008 -> 3

Okay I see. That helps, thank you!

Originally Posted by megafiddle

With &ptr as an argument, it will try to store the value 55 in ptr.
Now ptr, in all likelyhood, no longer points to b.

With *ptr as an argument, it will try to store the value 55 in whatever b is pointing to.
This is where the real havoc can be wrought.

-

Yeah I didn't expect for these statements to do anything specific, it was just a to get me thinking, and I have little experience with pointers. Thank you!