Sending variable size 2D array to function in C++

Question

I am trying to write a function which takes variable size array and prints it. But i am having problem in function declaration argument list while compiling.

The error is:

cannot convert ‘int (*)[(((sizetype)(((ssizetype)r) + -1)) + 1)]’ to ‘int**’ for argument ‘1’ to ‘void printHalf(int**, int, int)’
  printHalf( b, r, r);

Code:

#include <iostream>

using namespace std;

void printHalf( int **a, int row, int col ) {  // Problem is here.
    for( int i=0; i < row; i++) {
        for( int j=0; j < col; j++) {
            if( i <= j) {
                cout << a[i][j] << " ";
            }
        }
        cout << endl;
    }
}

int main() {
    int r;
    cout << "No. of rows: ";
    cin >> r;
    int b[r][r];

    for( int i=0; i < r; i++) {
        for( int j=0; j < r; j++) {
            cin >> b[i][j];
        }
    }

    printHalf( b, r, r);
    return 0;
}

What is causing this error, and how do i pass various arrays to function?

@coincoin I think you should use new[] instead of malloc() in C++. — MikeCAT
– MikeCAT, Commented Dec 9, 2015 at 15:39
I think you should use neither and use std::vector instead. — TartanLlama
– TartanLlama, Commented Dec 9, 2015 at 15:39

Simple · Accepted Answer · 2015-12-09 15:44:27Z

4

A 2D array of [N][M] is laid out the same in memory as a 1D array of [N*M].

void printHalf( int *a, int row, int col ) {  // Problem is here.
    for( int i=0; i < row; i++) {
        for( int j=0; j < col; j++) {
            if( i <= j) {
                cout << a[i*col+j] << " ";
            }
        }
        cout << endl;
    }
}

Then you can call it with printHalf( &b[0][0], r, r).

Your fundamental misunderstanding is the relationship between arrays and pointers. Arrays aren't pointers. An int** can be viewed as an array of int*, which isn't what you have. b[0] gets you an int[r]. This is different to an int*. b[0][0] gets you the first int.

edited Dec 9, 2015 at 15:44

answered Dec 9, 2015 at 15:42

Simple

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2 Comments

Pixelchemist Over a year ago

You'll want a[i*col+j]

Max Payne Over a year ago

Thank you for giving me C oriented answer, but i guess vector is the way to go for c++

R Sahu · Accepted Answer · 2015-12-09 15:44:29Z

3

Problems that I see

C++ does not allow declaration of variable length arrays. Hence,
```
int b[r][r];
```
is wrong.
A one dimensional array decays to a pointer but a two dimensional array does not decay to a pointer to pointer. Even if you had b defined correctly, such as:
```
int b[10][10];
```
you cannot use b as an argument to a function that expects an int**.

Solution

I suggest use of std::vector.

std::vector<std::vector<int>> b(r, std::vector<int>(r));

Change the function printHalf accordingly.

void printHalf( std::vector<std::vector<int>> const& b ) { ... }

You don't need row and col as arguments since that information can be obtained from the std::vector.

answered Dec 9, 2015 at 15:44

R Sahu

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Comments

NathanOliver · Accepted Answer · 2015-12-09 15:40:24Z

2

First off int b[r][r]; is a variable length array and is not standard in C++. If you need a container that can be sized at run time then I suggest you use a std::vector

std::vector<std::vector<int>> data(r, std::vector<int>(r, 0)); // create vector and set all elements to 0

Then your print function would become

void printHalf(const std::vector<std::vector<int>>& data) {
    for( std::size_t i=0; i < data.size(); i++) {
        for( std::size_t j=0; j < data[i].size(); j++) {
            if( i <= j) {
                cout << a[i][j] << " ";
            }
        }
        cout << endl;
    }
}

answered Dec 9, 2015 at 15:40

NathanOliver

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3 Comments

MikeCAT Over a year ago

Using reference to avoid copying many data might be good. void printHalf(std::vector<std::vector<int>> data) -> void printHalf(const std::vector<std::vector<int>>& data)

coincoin Over a year ago

I would also recommend to use a flatten/linear vector<int> only instead of nested vectors vector<vector<int>>.

NathanOliver Over a year ago

@coincoin That is a good suggestion and in production code that is what I would use. I did not want to present that to the OP as I am already changing to using a vector and wanted to leave the code looking as close the the original as I could.

Филя Усков · Accepted Answer · 2024-06-16 22:24:00Z

0

you can use extension/type Variable Length Array:

#include <stdio.h>
#include <stdlib.h>

int main() {
    int h,w;
    scanf("%d%d",&h,&w);
    int * arr = malloc(w*h*sizeof(int));
    for(int i=0; i<w*h; i++)
        arr[i] = i;

    typedef int (*arr2_t)[w];
    int (*arr2)[w];
    arr2 = (arr2_t)arr;

    for(int y=0; y<h; y++){
        for(int x=0; x<w; x++)
            printf("%d ",arr2[y][x]);
        printf("\n");
    }
}

I don't know how type can depend on variable, but gcc and clang compile that.

edited Jun 16, 2024 at 22:24

answered Jun 16, 2024 at 22:17

Филя Усков

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