pointers leaving me clueless, odd output, C?

Question

so i'm screwing around just to get a feel for pointers, and also to familiarize myself with malloc() and friends.
anyways here is my little program:

#include <stdio.h>

int in;
int *p1 = ∈

int main(void)
{
printf("enter decimal: ");
scanf("%d", p1);
printf("%d", p1);
return 0;
}

/*what ends up happening is i get random numbers 7 digits long...are they the addresses of memory in decimal? i know if i change the last printf to %p i just get the address of memory that the pointer points to, in hex. how can i get it to use the pointer to print out the decimal that was taken in?*/

BobberKnob · Accepted Answer

*p1 will dereference your pointer after you declared it. 

One of the reasons newcomers are confused by pointers, is the asterisk having kind of a double-meaning.

when you declare a pointer 
its:

int * p1;  

then call that variable by: 
p1;   //Address
 to dereference it:
*p1; //Value at the address of p1

Bill · Answer

scanf("%d", *p1);
printf("%d", *p1);

you want the contents of p1(*p1) not p1


note;  Yep, my bad, I had hit submt when irelized you didn't get an error message. scanf wants a pointer.

Ratchetr · Answer

Either of these will work:

printf("%d", in);
printf("%d", *p1);

Don't change your scanf as suggested by another....it's right the way it is.

pointers leaving me clueless, odd output, C?

3 Antworten