C++: Passing array of structs to functions by reference [duplicate]

Question

Please do not mark as duplicate, while there are similar questions to mine, they were either answered using pointers, did not address the same issue that I have, or they pertained to the language C instead of C++.

I am trying to pass an array of structs to a function, but when I try to do so, I am getting an error: declaration of ‘a’ as array of references. I think the problem may lay in the fact the compiler is reading it as an array instead of as a struct, but I am not sure how to remedy this problem.

I have defined a struct of three elements:

struct StructA {
   string name; 
   float income; 
   int amount; 
}

Declared this struct within main:

StructA a[15];

And am passing it to a function like so:

void FunctionA(StructA& a[], int& count) { //& to pass by reference

}

Furthermore, right below the definition of StructA, I have the matching function prototype for the function above.

What am I doing wrong?

EDIT: Someone marked this question as a duplicate of another; it is not. I am in an intro to programming class, and as I clearly stated up top, I cannot use a pointer. Guess what the other question uses? A pointer. See the difference? Furthermore, on a more practical level, I already tried the solution recommended in Pass a dynamic array of objects to function and it did not work for me and since I cannot comment in the other question, I have to ask a new question. I will just wait until I can ask my professor, grazie.

Is it a must to have to have an array? If not try declaring a std::vector<T> of StructA and pass a std::vector<StructA>& container to your function. — Francis Cugler
– Francis Cugler, Commented May 11, 2017 at 2:03

Weak to Enuma Elish · Accepted Answer · 2017-05-11 00:49:42Z

1

To pass an array of any size by reference, you can do:

template <size_t Size>
void FunctionA(StructA (&a)[Size])

It'll deduce the size of the array, so you don't need to specify it each time.

StructA a1[10];
StructA a2[15];

FunctionA(a1);
FunctionA(a2);

answered May 11, 2017 at 0:49

Weak to Enuma Elish

4,6493 gold badges26 silver badges36 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Shadi · Accepted Answer · 2017-05-11 15:24:05Z

1

Use this syntax:

void FunctionA(StructA (&a) [SIZE], int &count)

SIZE should be mentioned in this case.

Your syntax (void FunctionA(StructA &a [SIZE])) means that you pass array of references(&) not a reference to an array.

You can also use a template with the function:

template <size_t Size>
void FunctionA(StructA (&a) [SIZE])

You do not need to pass count if you mean it to be size of the array as noted by:Amadeus

Using template will enable the function for different sizes as answered by: James Root

You can use the function then as follows:

StructA a[10];
StructA b[20];

FunctionA(a);
FunctionA(b);

edited May 11, 2017 at 15:24

answered May 11, 2017 at 0:35

Shadi

1,7912 gold badges19 silver badges28 bronze badges

2 Comments

Matt Over a year ago

Used your method but I am getting a new error: parameter ‘a’ includes reference to array of unknown bound

Shadi Over a year ago

@Matt: you should use SIZE. I edit the answer and make it more detailed.

Collectives™ on Stack Overflow

C++: Passing array of structs to functions by reference [duplicate]

2 Answers 2

Comments

2 Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

2 Comments

Linked

Related