pointers... need reassurance I'm understanding things correctly

Hello guys, I've been working on a tutorial this morning and I believe I'm on the right track, I'm just looking for a 'yay or nay' as to if this makes sense or not.

here's my code...

Code:

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#include <stdio.h>


char mem[20];        // assume mem is at base address 2000
int x;
int val1 = 0      ;
int val2 = 1      ;
char *ptr;


int main (void) {
ptr = &mem[val1];    // get the address of array element val1 within mem


 printf ("base address (which can be considered as 2000) %d\n", mem); //my addition...prints initial base address
 
for (x = 0; x < 5; x++) {
    *ptr = 'A' + x + val2;
    printf ("address %d\n", ptr);  //my addition...this prints the address of the element you're currently at
    printf ("value %d\n", *ptr); //my addition...this prints what is contained inside that element
    ptr++;
}
getchar ();  //my addition...window won't dissappear until hitting 'enter'
return 0;
}

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I have added all the printf's in order to see what's going on.

the output I receive when executing is as follows:

Code:

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base address (which can be considered as 2000) 4202784
address 4202784
value 66
address 4202785
value 67
address 4202786
value 68
address 4202787
value 69
address 4202788
value 70

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I am being asked to state the 5 addresses that will be written to, and the value put into memory at that location.

assuming I've used 'printf' in the correct fashion,

can I then assume that my 5 addresses and values are:
2000 (value 66)
2001 (value 67)
2002 (value 68)
2003 (value 69)
2004 (value 70)

...(being that I can't set the base address, I let windows choose a different number to represent 2000)


Thanks in advance,

Just need to to verify what I'm doing makes sense!

You're trying to set a pointer's value directly, which isn't allowed (except for the special case of 0 or NULL)
ptr contains the address of a variable which holds a char.

so

Code:

---

char x = 'A'; /* x is the variable, it holds the character 'A' */
char *ptr; /* a pointer to it */

ptr = &x; /* make ptr point to x, it holds the address of x */
printf("address %p value %c\n", ptr, *ptr); /* print out ptr, and what it points to */
*ptr = 'F'; /* set what ptr point to to 'F' */
printf(""x now holds %c\n", x); /* the variable x has been updated, through ptr */

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My guess is that in your second printf you should use: "value %c\n" so that your output looks like 'B', 'C', 'D', 'E', 'F'. The rest of the program is fine.

A pointer is printed using %p in the printf family


You can get the address of the array by using &mem[0] (the address of the first element), or you are allowed to use mem on it's own (which gets assesed as &mem[0])


I would prefer to do both printf's at the same time, like Malcolm suggested. Also, I agree with nonoob, in that you probably wanted to see the characters, and not the number - Or better yet, you can see both

Code:

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    char apple[20];
    int banana;
    int i;


    char *pear;


    pear = &apple[0];


    printf ("base address (which can be considered as 2000) %p\n", apple);


    for (banana = 0, i = 0; i < 5; i++, pear++)
    {
        *pear = 'A' + i + banana;


        printf("\n\raddress %p \n\rCharacter '%c' \n\rASCII value %d \n", pear, *pear, *pear);


    }


    ...

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Quote:

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Originally Posted by Malcolm McLean

You're trying to set a pointer's value directly, which isn't allowed (except for the special case of 0 or NULL)

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That is not true.

Quote:

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Originally Posted by grumpy

That is not true.

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I thought so... I was confused by what he said, as my code worked perfectly.



...Thanks everyone for the clarification! Your responses were all very helpful