What you describe seems to be a common issue that many C or C++ developers can get (or have got) with sizeof.

Take a look at the following code:

char a[3];

int size(char b[]) {
    return sizeof b;
}

void setup() {
    int size1 = sizeof a;
    int size2 = size(a);
}

In that code, size1 will be evaluated, by the compiler, to 3 which is the real size of a variable, whereas size2 will be 2.

##Why so?

This is just because sizeof a is evaluated by the compiler, and the compiler knows that a is an array of 3 char; but when a is passed as an argument to function size(), it is converted to a pointer (to the first element of the array), hence its size is the size of a pointer, which is 2 bytes on AVR ATmega MCU.

This subtlety of sizeof may explain strange values as you describe.

##How to fix the issue?

Most C functions that take an array as argument will require another argument that tells the size of the array; that's the way (the only one I think) to deal with this problem.

Here is a sample code:

Servo allServos[] = {...};
#define NUM_SERVOS (sizeof(allServos) / sizeof(Servo))
...
void handleServos(Servo *servos, int numServos) {
    for (int i = 0; i < numServos; i++) {
        servos[i].write(...);
        ...
    }
}

void loop() {
    handleServos(allServos, NUM_SERVOS);
}

It is common practice in C to declare an array and immediately after, define its size as above.

What you describe seems to be a common issue that many C or C++ developers can get (or have got) with sizeof.

Take a look at the following code:

char a[3];

int size(char b[]) {
    return sizeof b;
}

void setup() {
    int size1 = sizeof a;
    int size2 = size(a);
}

In that code, size1 will be evaluated, by the compiler, to 3 which is the real size of a variable, whereas size2 will be 2.

Why so?

This is just because sizeof a is evaluated by the compiler, and the compiler knows that a is an array of 3 char; but when a is passed as an argument to function size(), it is converted to a pointer (to the first element of the array), hence its size is the size of a pointer, which is 2 bytes on AVR ATmega MCU.

This subtlety of sizeof may explain strange values as you describe.

How to fix the issue?

Most C functions that take an array as argument will require another argument that tells the size of the array; that's the way (the only one I think) to deal with this problem.

Here is a sample code:

Servo allServos[] = {...};
#define NUM_SERVOS (sizeof(allServos) / sizeof(Servo))
...
void handleServos(Servo *servos, int numServos) {
    for (int i = 0; i < numServos; i++) {
        servos[i].write(...);
        ...
    }
}

void loop() {
    handleServos(allServos, NUM_SERVOS);
}

It is common practice in C to declare an array and immediately after, define its size as above.

What you describe seems to be a common issue that many C or C++ developers can get (or have got) with sizeof.

Take a look at the following code:

char a[3];

int size(char b[]) {
    return sizeof b;
}

void setup() {
    int size1 = sizeof a;
    int size2 = size(a);
}

In that code, size1 will be evaluated, by the compiler, to 3 which is the real size of a variable, whereas size2 will be 2.

##Why so?

This is just because sizeof a is evaluated by the compiler, and the compiler knows that a is an array of 3 char; but when a is passed as an argument to function size(), it is converted to a pointer (to the first element of the array), hence its size is the size of a pointer, which is 2 bytes on AVR ATmega MCU.

This subtlety of sizeof may explain strange values as you describe.

What you describe seems to be a common issue that many C or C++ developers can get (or have got) with sizeof.

Take a look at the following code:

char a[3];

int size(char b[]) {
    return sizeof b;
}

void setup() {
    int size1 = sizeof a;
    int size2 = size(a);
}

In that code, size1 will be evaluated, by the compiler, to 3 which is the real size of a variable, whereas size2 will be 2.

##Why so?

This is just because sizeof a is evaluated by the compiler, and the compiler knows that a is an array of 3 char; but when a is passed as an argument to function size(), it is converted to a pointer (to the first element of the array), hence its size is the size of a pointer, which is 2 bytes on AVR ATmega MCU.

This subtlety of sizeof may explain strange values as you describe.

##How to fix the issue?

Most C functions that take an array as argument will require another argument that tells the size of the array; that's the way (the only one I think) to deal with this problem.

Here is a sample code:

Servo allServos[] = {...};
#define NUM_SERVOS (sizeof(allServos) / sizeof(Servo))
...
void handleServos(Servo *servos, int numServos) {
    for (int i = 0; i < numServos; i++) {
        servos[i].write(...);
        ...
    }
}

void loop() {
    handleServos(allServos, NUM_SERVOS);
}

It is common practice in C to declare an array and immediately after, define its size as above.