c++ error: invalid types 'int[int]' for array subscript

Question

Trying to learn C++ and working through a simple exercise on arrays.

Basically, I've created a multidimensional array and I want to create a function that prints out the values.

The commented for-loop within Main() works fine, but when I try to turn that for-loop into a function, it doesn't work and for the life of me, I cannot see why.

#include <iostream>
using namespace std;


void printArray(int theArray[], int numberOfRows, int numberOfColumns);

int main()
{

    int sally[2][3] = {{2,3,4},{8,9,10}};

    printArray(sally,2,3);

//    for(int rows = 0; rows < 2; rows++){
//        for(int columns = 0; columns < 3; columns++){
//            cout << sally[rows][columns] << " ";
//        }
//        cout << endl;
//    }

}

void printArray(int theArray[], int numberOfRows, int numberOfColumns){
    for(int x = 0; x < numberOfRows; x++){
        for(int y = 0; y < numberOfColumns; y++){
            cout << theArray[x][y] << " ";
        }
        cout << endl;
    }
}

Almost duplicate of c++ - invalid types 'int[int]' for array subscript - Stack Overflow -- although this question includes info how to pass 2D array to a function too. — user202729
– user202729, Commented Jan 28, 2021 at 4:01

Ben Voigt · Accepted Answer · 2011-05-29 02:19:52Z

7

C++ inherits its syntax from C, and tries hard to maintain backward compatibility where the syntax matches. So passing arrays works just like C: the length information is lost.

However, C++ does provide a way to automatically pass the length information, using a reference (no backward compatibility concerns, C has no references):

template<int numberOfRows, int numberOfColumns>
void printArray(int (&theArray)[numberOfRows][numberOfColumns])
{
    for(int x = 0; x < numberOfRows; x++){
        for(int y = 0; y < numberOfColumns; y++){
            cout << theArray[x][y] << " ";
        }
        cout << endl;
    }
}

Demonstration: http://ideone.com/MrYKz

Here's a variation that avoids the complicated array reference syntax: http://ideone.com/GVkxk

If the size is dynamic, you can't use either template version. You just need to know that C and C++ store array content in row-major order.

Code which works with variable size: http://ideone.com/kjHiR

edited May 29, 2011 at 2:19

answered May 29, 2011 at 2:08

Ben Voigt

286k45 gold badges444 silver badges764 bronze badges

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

4 Comments

Lightness Races in Orbit Over a year ago

To be pedantic, using references increases backward compatibility concerns, since C has no references :P

Ben Voigt Over a year ago

@Tomalek: Backward compatibility states that existing valid C code should still do the same thing when compiled as C++. This isn't valid C code, so there's no requirement to behave the same way as C.

corlettk Over a year ago

That printArray(*sally, 2, 3); implementation is what I was alluding to... but I got it completely wrong! Thanx for the fish.

Ben Voigt Over a year ago

@corlettk: Oh, ok.... I thought you meant jagged array style, and I read your comment twice. The version with *sally does require the dimensions (except the first) to be known at compile time, they're just hidden in the type T inferred by the compiler.

Frédéric Hamidi · Accepted Answer · 2011-05-29 01:48:06Z

2

Since theArray is multidimensional, you should specify the bounds of all its dimensions in the function prototype (except the first one):

void printArray(int theArray[][3], int numberOfRows, int numberOfColumns);

answered May 29, 2011 at 1:48

Frédéric Hamidi

264k42 gold badges497 silver badges486 bronze badges

3 Comments

corlettk Over a year ago

... or pass the matrix as an array of pointers-to-int-array... which is one alternative when you don't know (at "write time", ie when you are writing the printArray method) what the subsequent-dimension(s) are. Another alternative is to #define the subsequent-dimension(s). I prefer the latter approach, where possible.

Ben Voigt Over a year ago

@corlettk: A C++ (or C) multidimensional array is not an array of pointers.

Ben Voigt Over a year ago

@corlettk: Which is not to say you couldn't do things that way, you'd just have to set up the pointer array first.

wiredolphin · Accepted Answer · 2020-02-23 17:15:34Z

1

I'm aware of the date of this post, but just for completeness and perhaps for future reference, the following is another solution. Although C++ offers many standard-library facilities (see std::vector or std::array) that makes programmer life easier in cases like this compared to the built-in array intrinsic low-level concepts, if you need anyway to call your printArray like so:

printArray(sally, 2, 3);

you may redefine the function this way:

void printArray(int* theArray, int numberOfRows, int numberOfColumns){
    for(int x = 0; x < numberOfRows; x++){
        for(int y = 0; y < numberOfColumns; y++){
            cout << theArray[x * numberOfColumns + y] << " ";
        }
        cout << endl;
    }
}

In particular, note the first argument and the subscript operation:

the function takes a pointer, so you pass the name of the multidimensional array which also is the address to its first element.
within the subscript operation (theArray[x * numberOfColumns + y]) we access the sequential element thinking about the multidimensional array as an unique row array.

edited Feb 23, 2020 at 17:15

answered Sep 18, 2017 at 9:03

wiredolphin

1,6511 gold badge20 silver badges27 bronze badges

3 Comments

Cătălina Sîrbu Over a year ago

Exactly what i was searching for. Could you explain a little bit more the way you computed the formula between brackets?

wiredolphin Over a year ago

@Cătălina Sîrbu, as stated by the second point of the ending list, the formula returns the integer of the current element if you traverse the array thinking to it as a one dimensional array. If that is not yet clear, you should before understand how arrays, and particularly multidimensional arrays, are effectively stored in memory. Tell me if you need further explanations.

Cătălina Sîrbu Over a year ago

i got it finnaly

neuront · Accepted Answer · 2011-05-29 01:50:52Z

0

If you pass array as argument you must specify the size of dimensions except for the first dim. Compiler needs those to calculate the offset of each element in the array. Say you may let printArray like

void printArray(int theArray[][3], int numberOfRows, int numberOfColumns){
    for(int x = 0; x < numberOfRows; x++){
        for(int y = 0; y < numberOfColumns; y++){
            cout << theArray[x][y] << " ";
        }
        cout << endl;
    }
}

answered May 29, 2011 at 1:50

neuront

9,6426 gold badges46 silver badges74 bronze badges

Collectives™ on Stack Overflow

c++ error: invalid types 'int[int]' for array subscript

4 Answers 4

4 Comments

3 Comments

3 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

4 Comments

3 Comments

3 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related