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Related

function in C++

billpenner
billpenner over 10 years ago

When a function is finished, closed bracket}, where does the programming go?

My programming seems to fall through, that is the next function after the current function is complete seems to be running. Are incorrect brackets  causing my problem?

For example:

void loop() ;

{

do my first thing;

if test 1 true, go to void thing2

if test 2 true, go to void thing3

}

Where am I at this point? Am I back in the loop or do I fall through thing3 to what is after it in the program.

Please straighten me out a little. I am new to C++.

Thanks

Bill

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Top Replies

  • Robert Peter Oakes
    Robert Peter Oakes over 10 years ago in reply to gadget.iom +1 suggested
    OK, Call out the censors, he used the GOTO word…….. Ok, more seriously, GOTO is a bad thing to use and can leave you with a mess from a memory and cpu resource perspective, most modern languages do not…
  • mcb1
    mcb1 over 10 years ago in reply to Robert Peter Oakes +1
    GOTO is a bad thing to use and can leave you with a mess I know its disliked by some programmers, but I've seen others say there is nothing wrong with its use.... I guess there will always be opposing…
Parents
  • gadget.iom
    0 gadget.iom over 10 years ago

    Hi billpenner

     

    The loop function will repeat indefinitely.

    The go-to parts of your code could transfer the program flow to the associated labels, from which the code will continue, essentially skipping any code in

    between.

    eg:

    for(byte r = 0; r < 255; r++){
      for(byte g = 255; g > -1; g--){
      for(byte b = 0; b < 255; b++){
      if (analogRead(0) > 250){ goto bailout;}
      // more statements ...
      }
      }
    }
    bailout:

     

     

     

    A better approach may be to turn the goto's into function calls. in this case the program flow will move to the function you declared and then return back to the point immediately after the function call.

    eg.

    void setup(){
      Serial.begin(9600);
    }
    
    void loop() {
      int i = 2;
      int j = 3;
      int k;
    
      k = myMultiplyFunction(i, j); // k now contains 6
      Serial.println(k);
      delay(500);
    }
    
    int myMultiplyFunction(int x, int y){
      int result;
      result = x * y;
      return result;
    }

     

    More information can be found on the following pages:

    Arduino - Goto

    Arduino - FunctionDeclaration

     

     

    Does this clarify the situation somewhat?

     

    Paul

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  • billpenner
    0 billpenner over 10 years ago in reply to gadget.iom

    Yes it does.  But you bring up another confusing thing.

    When you say k = myMultiplyFunction(i, j); does that mean k  = a call to a function named " myMultiplyFunction")

    and you are taking j and k to that function?

    if so, why does all three variables ( myMultiplyFunction and I and j have to be declared as int again?

    I think I understand why 'result' has to be created so the answer can be returned but  could you just make

    k  = x * y

    return ?

    and is return necessary since the" } " ends the call function?

    These questions and you guys response are like having an instructor at my side. I have understood more today than I have with many hours studying C++.

    I can't thank you all enough.

    Bill

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  • gadget.iom
    0 gadget.iom over 10 years ago in reply to billpenner

    In that example 'myMultiplyFunction' is called and 'i' and 'j' are passed in as parameters. The value returned from the function is then loaded into k, and it's the return keyword that does this.

    If the function does not need to return data the keyword can be omitted.

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  • gadget.iom
    0 gadget.iom over 10 years ago in reply to billpenner

    In that example 'myMultiplyFunction' is called and 'i' and 'j' are passed in as parameters. The value returned from the function is then loaded into k, and it's the return keyword that does this.

    If the function does not need to return data the keyword can be omitted.

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