CODEWITHSUNDEEP
c++arraysfunctionmultidimensional-arraycompiler-errors
Max Payne
Max Payne

Reputation: 2428

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

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?

Upvotes: 3

Views: 2782

Answers (4)

Филя Усков
Филя Усков

Reputation: 302

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.

Upvotes: 0

R Sahu
R Sahu

Reputation: 206607

Problems that I see

  1. C++ does not allow declaration of variable length arrays. Hence, 

    int b[r][r];

    is wrong.

  2. 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.

Upvotes: 3

Simple
Simple

Reputation: 14390

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.

Upvotes: 4

NathanOliver
NathanOliver

Reputation: 180630

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;
    }
}

Upvotes: 2

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