学习C - 指针和内存寻址

Question

I am learning C programming and I have a simple question about pointers...

I used the following code to play around with pointers:

#include <stdio.h>

int main (int argc, const char * argv[]) {

int * c;

printf("%x\n",c);
return 0;
}

When I print the value of C, I get back a 0. However, when I print &c (ie printf("&x\\n",&c) I get an address in memory...

Shouldn't I be getting an address in memory when printing the pointer (ie printf("%x\\n",c)?

--- EDIT ---

#include <stdio.h>
#include <stdlib.h>

int main (int argc, const char * argv[]) {

    char * a = malloc(11);
    printf("String please\n");
    scanf("%s",a);
    printf("%s",a);

}

The question is, why does printf("%s",a) returns the string instead of the address that is stored in a?

Shouldn't I use *a to follow the pointer and then print the string?

Answer 1

your current program is not correct. You define variable and do not set value before first use. the initial value is not guranteed for c , but you are lucky and it is equal to 0 . It means that c points to nowhere. when you print &c you print address of variable c itself. So actually both versions print address.

Answer 2

printf is actually quite a complex function and can be made to do all sorts of tricks by giving it the right format specifier.

In your string example:

printf("%s", a)

the "%s" tells the printf function that the following variable should be treated as a string. In C, a string is a pointer to one or more char, terminated by a char containing 0. This is a pretty common request, which is why printf supports a format specifier "%s" that triggers this relatively complex behavior. If you want to print the address contained in the string pointer you have to use the format you found earlier:

printf("%x\n",a);

which tells printf to treat the contents of a as an unsigned integer, and print it in base 16.

*a would just be the value pointed to by a, which is just a single character.

You could print the single character with

printf("%c", *a);

Answer 3

Having int* c; If you print value of c, you get back a value that should be interpreted as a memory address of an integer value. In you example it might be 0 or something completely different as you are not initializing c.

If you print &c you get memory address of the pointer c (stored in stack).

Answer 4

#include <stdio.h>

int main (int argc, const char * argv[]) {

int * c;
int a=10;
c = &a;

printf("%x\n",c);
return 0;
}

This may clarify what happens when you make the int pointer point to something in memory.