How to manipulate a variable value using an array?

[amateurrobotguy]

I really really really want to be able to use an array to call my relays. How can i do this? See below post for what i want to be able to do.

[amateurrobotguy]

Better yet, how can a store variable names within an array so that when i say:
myVar = 1;
array[1] = {myVar};

array[1] = 2;
printf(myVar);

I want the displayed output to be 2

[Phil Mack]

Originally Posted by amateurrobotguy

how can a store variable names within an array

You can store the address of the variable in the array, then use that address as a pointer to the value you want to print. Your array would be declared as an array of pointers.

myVar = 1;
array[1] = &myVar;

(*)(array[1]) = 2;
printf("%d\n",myVar);

Good luck,
~Phil

[Tom Bottiglieri]

Or use a macro.

#define MYVAR        array[1]

print(MYVAR);

[amateurrobotguy]

I need you to confirm if what I am thinking is correct:

myVar = 1; Establish the "shell" as being = 1
array[1] = &myVar; Let the 1st value of array[] = the address of the "shell"

(*)(array[1]) = 2; the * means modify/check the contents of the address of myVar
printf(*myVar); since the shell wasn't updated, the * is needed to display the contents of the address of myVar since printf(myVar) would return 1

[DanDon]

Originally Posted by amateurrobotguy

I need you to confirm if what I am thinking is correct:

myVar = 1; Establish the "shell" as being = 1
array[1] = &myVar; Let the 1st value of array[] = the address of the "shell"

(*)(array[1]) = 2; the * means modify/check the contents of the address of myVar
printf(*myVar); since the shell wasn't updated, the * is needed to display the contents of the address of myVar since printf(myVar) would return 1

 (*)(array[1]) = 2;

That would put a '2' in the memory cell pointed to by the address in array[1]. Or in this case....since the address in array[1] is that of myVar, the value of myVar would be 2, so assuming myVar is an int, your code for printing it out would be:

printf("%d\n",myVar);

Which would print out '2'.

[amateurrobotguy]

int *ptr;
int a, b;

a = 10;
ptr = &a;
b = *ptr;

Why bother with the int *ptr declaration when it has to be used again in the b= part? Is this just proper etiquette and not really needed?

Is this alright for my declaration:
int joystickbuttons[8] = {&p1_sw_trig, &p1_sw_top, &pl_sw_aux1, &p1_sw_aux2, &p2_sw_trig, &p2_sw_top, &p2_sw_aux1, &p2_sw_aux2};

And this for my looped search:
if ((*)joystickbuttons[arrayindexer[1]] == 1)

arrayindexer is just a simple array I am using for a counter.


If all this looks good then it means I finally figured this whole pointer thingie out largely thanks to this forum and Wiki.

[Tom Bottiglieri]

Originally Posted by amateurrobotguy

Why bother with the int *ptr declaration when it has to be used again in the b= part? This is the example from Wiki.

int *ptr; //ptr is a pointer to the address an 'int' somewhere in memory
int a, b; //compiler chooses where these go in memory, each have an address

a = 10; //set value of a to 10
//pass a by reference to ptr
ptr = &a; //ptr now holds the address of where a is in memory
b = *ptr; //b now holds the value (use *) that ptr points to

In this case, b would be equal to a at the end.
The first line declares a pointer, which is different from a regular variable.
You can do other cool things with this, too.

void idouble(int * input){
   *input *= 2 ;
}

void main(){
    int x ;
    x = 5 ;
    idouble(&x) ;
    print(x);
}

This will print 10.

[amateurrobotguy]

So i have to say that my array of buttons is going to be holding addresses by declaring it like this:

int (*)joystickbuttons[8] = {&p1_sw_trig, &p1_sw_top, &pl_sw_aux1, &p1_sw_aux2, &p2_sw_trig, &p2_sw_top, &p2_sw_aux1, &p2_sw_aux2};

And thus the * in the declaration indicates that the variable is "special" in that it will indeed be holiding addresses and not just numbers and crap.

And the 2nd half of the above code:

if ((*)joystickbuttons[arrayindexer[1]] == 1)

Will my above 2 pieces of code work?

I don't get some of your example:

void idouble(int * input){  Input the address and convert it to the data "input"
   *input += input ; I don't get the use of *input
}

void main(){
    int x ;
    x = 5 ;
    idouble(&x) ;  Call the above function giving it an address as a parameter
    print(x); I don't get how the idouble() returns x
}

[Tom Bottiglieri]

Originally Posted by amateurrobotguy

I don't get some of your example:

void idouble(int * input){  Input the address and convert it to the data "input"
   *input *= 2 ; I don't get the use of *input
}

void main(){
    int x ;
    x = 5 ;
    idouble(&x) ;  Call the above function giving it an address as a parameter
    print(x); I don't get how the idouble() returns x
}

input is a pointer. it holds an address. in this particular case, it holds the address of x. putting a * in front of a pointer means that you want to look at the value held at the address of the pointer. if input = &x, *input = x. so modifying *input will actually be modifying the value held in x. (input is a pointer, so it has no real value attached to it.) By doing this, you can modify variables in your main function without returning ANYTHING from a called function. idouble returns nothing, but because you are telling it where x is in memory and not just the value of x, you can actually change x from another function. Make sense?

[amateurrobotguy]

Oh you sly bastard You are basically bypassing the whole naming system of variables with input and directly puting data to addresses. I like this The int *input is just the declaration of input and not the actual conversion of &x to data....that was the confusing part. I can see why google refers to pointers as "advanced c".

Can you check the code i posted above the above post to tell me if that will work fine?