How to access the value of a local variable in a function, outside of it?

Question

When I run my code, It says "velocity" was not declared in the scope. I tried to add, at the top, a declaration for velocity equal to zero and the code works but then velocity never updates. How can I make it so i'm able to call velocity in my loop function? Code:

const int trigPin = 9;
const int echoPin = 10;
int trials = 0;
bool win = true;

long duration1, distance1, duration2, distance2;

void setup() {
  pinMode(trigPin, OUTPUT);
  pinMode(echoPin, INPUT);
  Serial.begin(9600);
}

void getDistance(){
  digitalWrite(trigPin, LOW);
  delayMicroseconds(2);
  digitalWrite(trigPin, HIGH);
  delayMicroseconds(10);
  digitalWrite(trigPin, LOW);

  long duration1 = pulseIn(echoPin, HIGH);
  long distance1 = (duration1*.0343)/2;

  Serial.print("Distance: ");
  Serial.println(distance1);
  delay(100);

  digitalWrite(trigPin, LOW);
  delayMicroseconds(2);
  digitalWrite(trigPin, HIGH);
  delayMicroseconds(10);
  digitalWrite(trigPin, LOW);

  long duration2 = pulseIn(echoPin, HIGH);
  long distance2 = (duration2*.0343)/2;

  Serial.print("Distance: ");
  Serial.println(distance2);
  delay(100);

  }


void calculateSpeed(long dist2,long dist1,long time2,long time1){
  long numerator = dist2 - dist1;
  long denominator = time2 - time1;
  long velocity = numerator / denominator;
  return velocity;
}

void loop() {
  while (win == true){
    getDistance();
    calculateSpeed(distance2, distance1, duration2, duration1);
    Serial.println(velocity);

}
}

Answer 1

Answer to the general question

You cannot access a local variable in a function from outside the function, however, there are multiple ways you can get a value from within a function. Here are a few examples. Note that I am using an int variable in the examples, but this works with other types as well such as long .

1. Return the value from the function:

int foo()
{
    int val = 42;
    return val;
}

int main()
{
    // Gets the value from foo()
    int val = foo();
}

2. Create a global variable and modify it within the function (not recommended)

There are multiple reasons why this is not good practice, but I'm not going to get into that here. Just ask Google why global variables are generally not good and you'll find a decent explanation.

// Global variable
int val;

void foo()
{
    val = 42;
}

int main()
{
    // val is accessible from here and is equal to 42
}

3. Pass the variable to the function by pointer and modify it within the function

void foo(int* val)
{
    *val = 42;
}

int main()
{
    int val;
    foo(&val);

    // val now contains 42
}

4. Pass the variable by reference and assign a value in the function

void foo(int& val)
{
    val = 42;
}

int main()
{
     int val;
     foo(val);

     // val now equals 42
}

---

Your specific issue:

You are trying to return a value from a void function. Change calculateSpeed to return long , and then use the returned value in your other function:

// Return long
long calculateSpeed(long dist2,long dist1,long time2,long time1){
  long numerator = dist2 - dist1;
  long denominator = time2 - time1;
  long velocity = numerator / denominator;
  return velocity;
}

void loop() {
  while (win == true){
    getDistance();

    // Get the returned value
    long velocity = calculateSpeed(distance2, distance1, duration2, duration1);

    // Use the returned value
    Serial.println(velocity);
  }
}

There are also a couple other issues though. You declare some global variables with long duration1, distance1, duration2, distance2; , but then you redeclare these variables locally within functions.

For example, in getDistance() you need to change this:

long duration1 = pulseIn(echoPin, HIGH);
long distance1 = (duration1*.0343)/2;

to this:

duration1 = pulseIn(echoPin, HIGH);
distance1 = (duration1*.0343)/2;

Answer 2

You cannot use the name of the local variable as visible only inside calculateSpeed .

But because you return it from that function you can do what return values are made for:

Serial.println(calculateSpeed(distance2, distance1, duration2, duration1));

As scohe001 kindly pointed out in a comment (thanks),
you need to match the function head to the fact that your return something:

long calculateSpeed(long dist2,long dist1,long time2,long time1){

Alternatively, also do the return value fix and then introduce a variable inside the calling function, fill it with the return value and then use it. But that is only needed if you want to continue using the value.

I wanted to show the code for that, too. But tjwrona1992s answer already covers that.