Using Char Array with pinMode [SOLVED]

Projects Programming
1

Hi,

How do I use Char Arrays with pinMode?

I did the following:

Assigned the output pins;

#define RELAY1  4                        
#define RELAY2  5                        
#define RELAY3  6                        
#define RELAY4  7
#define RELAY5  8                        
#define RELAY6  9                        
#define RELAY7  10                        
#define RELAY8  11

The array;

const char *relays[] = {"RELAY1","RELAY2","RELAY3","RELAY4","RELAY5","RELAY6","RELAY7","RELAY8"};

Set the output pins to LOW to check response;

  for (int i=0; i = 7; i++){
    pinMode(relays[i],OUTPUT);
    digitalWrite(relays[i],LOW); 
  }

and I get no response... If I use;

  pinMode(RELAY1,OUTPUT);
  digitalWrite(RELAY1,LOW);
  etc.....

The relays respond fine.

Thanks

Pete

2

does it make a difference if you change:

const char *relays[] = {"RELAY1","RELAY2","RELAY3","RELAY4","RELAY5","RELAY6","RELAY7","RELAY8"};

into:

const char* relays[] = {"RELAY1","RELAY2","RELAY3","RELAY4","RELAY5","RELAY6","RELAY7","RELAY8"};

(i have changed the "*" so it is after char, and not before the array name)

3

try

const char relays[] = { RELAY1, RELAY2, RELAY3, RELAY4, RELAY5, RELAY6, RELAY7, RELAY8 };
4

#define names are visible to the compiler at compile-time, but don't exist at
runtime. And names are not the same thing as strings (which are runtime
objects).

You need an array of int or byte:

#define RELAY1  4                        
#define RELAY2  5                        
#define RELAY3  6                        
#define RELAY4  7
#define RELAY5  8                        
#define RELAY6  9                        
#define RELAY7  10                        
#define RELAY8  11

byte relays[] = {RELAY1, RELAY2, RELAY3, RELAY4, RELAY5, RELAY6, RELAY7, RELAY8};

void setup ()
{
  for (int i=0; i = 7; i++){
    pinMode (relays[i], OUTPUT);
    digitalWrite (relays[i], LOW); 
  }
  ..
}

Whitespace serves only to separate names, it is ignored:

const char *relays[]

parses the same as

const char* relays[]
5

thanks MarkT

It works a treat....

Thanks everyone else for their input too

6

parses the same as

To the compiler. Not to me.

char *a, b;

looks to the compiler, and to me, that a is a pointer to char and that b is a char.

char* a, b;

looks to the compiler like a is a pointer to char and b is a char, but looks to me like a and b are pointers to char.

I always put the * with the variable. I never put it with the type.

7

problem is solved by using one declaration per line. No mistakes like this

char a, *b, **c, d[], *e[];

(OK I am exaggerating)

8

problem is solved by using one declaration per line.

Yes, it is.