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

This is a discussion on pointers... need reassurance I'm understanding things correctly within the C Programming forums, part of the General Programming Boards category; Hello guys, I've been working on a tutorial this morning and I believe I'm on the right track, I'm just ...

  1. #1
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    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:
    #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;
    }
    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:
    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
    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!

  2. #2
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    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 */
    I'm the author of MiniBasic: How to write a script interpreter and Basic Algorithms
    Visit my website for lots of associated C programming resources.
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  3. #3
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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.

  4. #4
    TEIAM - problem solved
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    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:
    
        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);
    
    
        }
    
    
        ...
    Fact - Beethoven wrote his first symphony in C

  5. #5
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    Quote Originally Posted by Malcolm McLean View Post
    You're trying to set a pointer's value directly, which isn't allowed (except for the special case of 0 or NULL)
    That is not true.
    Right 98% of the time, and don't care about the other 3%.

    If I seem grumpy or unhelpful in reply to you, or tell you you need to demonstrate more effort before you can expect help, it is likely you deserve it. Suck it up, Sunshine, and read this, this, and this before posting again.

  6. #6
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    Quote Originally Posted by grumpy View Post
    That is not true.
    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

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