How can we declare the function with call by reference array as argument?

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I am confused how I can declare and call the function with call by reference array as argument. I wrote a code and I got an error for declaration of the function about call by reference array. How can I solve that.

This is a declaration of the function:

C++

void lenEqualize(int &updDisStr2[], int shortLenStr[], int diffLen, int updLen);
// updDisStr2: is an array which should be called by reference
// shortLenStr: call by value array

call the function:

C++

lenEqualize(updDisStr2[], equiDisStr2[], diffLen, updLen)

Posted 4-Aug-15 22:21pm

amin.j

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Comments

Afzaal Ahmad Zeeshan 5-Aug-15 4:39am

Why are you even appending the indexers? Remove them.

amin.j 5-Aug-15 4:44am

I removed but it is not solved.

[no name] 5-Aug-15 4:50am

This is more complex than your example shows. We need all your code. Why do you wish to pass explicitly by reference? Only arrays of known size can be passed by reference in this manner. Is your array of known size? eg int marray[10];

amin.j 5-Aug-15 4:52am

You got that. It needs size. But the size is not specific at the start of program. Is it possible to have a vector with call by reference?

3 solutions

Solution 3

You need to uses template and add missing parenthesis if you want to pass reference and have the size and the caller must have array with known size.

For flexibility, you should use std::vector instead.

By the way, you cannot use the reference syntax if you want to create a new array or chane its size inside the function.

Although it can be done with pointers using something like Solution 2, I would recommand to use STL containers like vector instead. It is much more maintainable as the size and the data are kept together and would always match.

Posted 5-Aug-15 7:11am

Philippe Mori

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CPallini · Accepted Answer · 2015-08-04T22:41:00

Solution 1

Quote:
void lenEqualize(int &updDisStr2[], int shortLenStr[], int diffLen, int updLen);

First argument is an array of references to int values ( the [] operator binds tightly than the & one, see "C++ operator precedence"[^]). Anyway I don't get your point why should you pass an array by reference?

Posted 4-Aug-15 22:41pm

CPallini

Comments

amin.j 5-Aug-15 4:49am

I want to change the values of that array in the function. And don't return sth.

CPallini 5-Aug-15 5:06am

You already can: there's no need to pass it by reference. C/C++ pass the array 'name' that is a pointer to the first item by value that prevents the modification of the pointer value itself NOT of the pointed elements (the array items).

try
void fun( int a[])
{
a[1] = 42;
}
int main()
{
int b[] = {1,2,3};
fun(b);
std::cout << b[1] << std::endl;
}

Richard MacCutchan · Accepted Answer · 2015-08-04T22:43:00

Solution 2

You seem to be confused about how to reference arrays. An array declared as int myints[5] can be referenced just by its name alone, no prefix ampersands and no suffix brackets. I'm not sure what you mean by call by value array.
So your code should be something like:

C++

void lenEqualize(int* updDisStr2, int* shortLenStr, int diffLen, int updLen)
{
    // updDisStr2: is a pointer to an array
    // shortLenStr: is also a pointer to an array
}
//call the function:
int updDisStr2[10];
int shortLenStr[10];
int diffLen;
int updLen;
lenEqualize(updDisStr2, equiDisStr2, diffLen, updLen)</p>

Note that the names declared in the function parameters are local to that function and do not need to match the originals.

Posted 4-Aug-15 22:43pm

Richard MacCutchan

Updated 4-Aug-15 22:44pm

v2

Comments

amin.j 5-Aug-15 5:12am

I corrected the code by your comments, the error is solved. Now, how can I reach the element of updDisStr2? I used this for the first element of this array:

cout << *updDisStr2[0] ;

but it sends error.

Richard MacCutchan 5-Aug-15 5:18am

Yes because you are trying to dereference something that is not a pointer. You should use either:
cout << *updDisStr2; // the element that updDisStr2 is pointing to
or
cout << updDisStr2[0]; // the element at offset 0 of updDisStr2
Which, in this case are both the same. You can iterate through the elements by using the name as a pointer


for (int i = 0; i < MAXVALUE; ++i)
{
    cout << *updDisStr2++; // increment the pointer after writing the current value
}

or use explicit offset values.


for (int i = 0; i < MAXVALUE; ++i)
{
    cout << updDisStr2[i]; // i is the current offset
}

amin.j 5-Aug-15 5:29am

This is part of my code that gives error: when I changed that to
cout << updDisStr2[0];
now it say that updDisStr2 is not declared, but I declared it in the if. I am really confused?

if (lenStr1 > lenStr2){
updLen = lenStr1;
int updDisStr2[updLen];
lenEqualize(updDisStr2, equiDisStr2, diffLen, updLen);
}
else if (lenStr1 < lenStr2){
updLen = lenStr2;
int updDisStr1[updLen];
lenEqualize(updDisStr1, equiDisStr1, diffLen, updLen);
}

Richard MacCutchan 5-Aug-15 7:21am

You need to show us the code for the lenEqualize function. Try and collect all the relevant information together so we can see exactly what you are doing.

Richard MacCutchan 5-Aug-15 7:53am

As I suggested in your other question. You really need to get hold of a good C++ book and study the basics. The difference between an element and an array, and the use of pointers versus indexers is something that you really need to understand fully. Also you need to understand scope and lifetime of variables. You cannot safely create a local array inside a function and return it to the caller by address.

amin.j 5-Aug-15 7:59am

I read book and know basics of that. But when it comes to practical, it is a different story. First, I didn't write the code in this way which I defined array outside function, but some people said doing this way. Now, what is your suggestion: there is two ways, first I define an array and give it to the function to change, second define the array inside the function and return it.

Richard MacCutchan 5-Aug-15 8:13am

I'm sorry, but you still have a way to go. Whatever you read in that book you still do not fully understand the use of arrays and pointers, and scope and lifetime. I know that these are not easy subjects, but until you clearly understand them you are going to struggle even more.

As to your last question, it is better to create the array outside the function, and let the function change it. However, that is really the old style C code. For C++ you should be using one of the built in classes that handles collections of data. See https://msdn.microsoft.com/en-us/library/cscc687y(v=vs.100).aspx. And yes, I know it is more to learn, but if this is your chosen career path, it will pay dividends in the long term.

amin.j 5-Aug-15 13:16pm

Dear Richard, Many thanks for your suggestions. I did that and it is solved. i have a question: Do you know some tutorial or sth that really teaches great features of C++? I mean to read sth useful in a short time.

Richard MacCutchan 6-Aug-15 3:51am

Sorry, but I have not looked at a learning book for many years. I can only suggest you go to https://www.google.com/search?q=c%2B%2B%20books, and look at some of the comments by other readers.