Invalid operands to binary *

This is a discussion on Invalid operands to binary * within the C Programming forums, part of the General Programming Boards category; This is probably a pretty simple problem to fix, but it is causing me a great deal of concern. The ...

  1. #1
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    Question Invalid operands to binary *

    This is probably a pretty simple problem to fix, but it is causing me a great deal of concern. The problem appears in the line
    Code:
    spot = (point+place)*y;
    Where spot is an integer pointer, point is another integer pointer, place is an integer, as is y. I definitely need to add place and point before multiplication, but it just wont accept my ideas. Much like society at large...

  2. #2
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    If spot and point are pointers, you need to dereference them before doing your calculations with them.

    The actual value of spot is a memory address like 00430030.
    Code:
    #include <stdio.h>
    #include <stdlib.h>
    
    int main()
    {
    	int *pointer = malloc(sizeof(*pointer));
    
    	*pointer = 5;
    
    	printf("pointer = &#37;d\n", pointer);
    	printf("*pointer = %d\n", *pointer);
    	
    	return 0;
    }
    Output:
    Code:
    pointer = 4390960
    *pointer = 5
    Don't quote me on that... ...seriously

  3. #3
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    I'm not sure if that's going to work for my particular problem
    point is the beginning of a dynamically allocated array, and as such I don't really want to change it, or its contents.

  4. #4
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    Quote Originally Posted by jrfall349 View Post
    I'm not sure if that's going to work for my particular problem
    point is the beginning of a dynamically allocated array, and as such I don't really want to change it, or its contents.
    You cannot add pointer to pointer. You can add int to pointer to get a new address or add value stored in the pointed location to another value. to do so - you should dereference your pointer:

    Code:
    *point + *place
    The first 90% of a project takes 90% of the time,
    the last 10% takes the other 90% of the time.

  5. #5
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    Dereferencing by itself does not change anything. It just accesses the data that the pointer points to. And if point is an array then wouldn't you want to be accessing elements of that array rather than the memory address? Like:
    Code:
    point[i]
    Don't quote me on that... ...seriously

  6. #6
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    Guess I'm just not sure what dereferencing is actually doing. We haven't covered that yet.

  7. #7
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    Code:
    Memory
    Address		Value
    ==========================
    		|--------|
    0x458700	|0x458704| <- pretend this is the pointer
    		|--------|
    0x458704	|00000015|
    		|--------|
    0x458708	|00010555|
    		|--------|
    So then if you have a pointer, it's value may be 0x458704. int *pointer; pointer would be 0x45804. If you do something to it like pointer++, pointer would be 0x45808. That's not what you want. You want to change the value that pointer is pointing to, 15 in this case. '*' dereferences the pointer. *pointer is equal to 15 because 15 is the value stored at memory address 0x458704. (*pointer)++ would change the 15 to 16. The value at memory address 0x458704, would now be 16. The pointer is still equal to 0x458704.
    Don't quote me on that... ...seriously

  8. #8
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    Okay, I think I get it... I don't want to add to point, I want to add to the value of point...
    Only problem I get now is

    [Warning] assignment makes pointer from integer without a cast

    And my program crashes. Urg this thing's starting to bug me.

  9. #9
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    If the code isn't extremely large, post it all.
    Don't quote me on that... ...seriously

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