Passing a two-dimensional array as a function argument

Question

I can not figure out the transfer of a two-dimensional array to a function, and its output by this function on the screen. And what other ways can you pass a two-dimensional array to a function?

#include <stdio.h> int printmas (int intmas[3][4]) { for(int i=0; i<3; i++) { for(int j=0; j<4; j++) { printf("%d ", intmas[i][j]); } printf("\n"); } } int main () { int danmassiv [3][4]={{10,9,98,65}, {8,-9,-4,6}, {15,6,78,-8}}; printmas(danmassiv); }

does not specifically output the given array and I cannot figure out how to send it to the function
Everything is perfectly going, working, transmitting and outputting.
By the way, on the page with a question (that is, the same), on the right, there is a list of related questions, under the heading "Related".
If you are given an exhaustive answer, mark it as correct (a daw opposite the selected answer).

Yaroslav yaroslav 2,495 one 6 17 · Accepted Answer · 2016-06-05T16:25:45

The argument function is a two-dimensional array.

 int fun(int array[][]) int fun(int * array[]) int fun(int ** array)

avp avp 37.4k 3 35 90 · Answer 2 · 2016-06-05T17:05:37

In the general case, it may be best to pass the dimensions of the array and the address of its first element to the function (that is, to represent this array as one-dimensional in the function) and independently calculate the required index when accessing the elements of the array.

For example, for a two-dimensional array:

 void pri_arr (int ncol, int nlines, int a[]) { int i, j; for(i = 0; i < nlines; i++) for (j = 0; j < ncol || (puts(""), 0); j++) printf("%d ", a[i * ncol + j]); } int main() { puts("2 dim"); int a[3][4] = {{1, 2, 3, 4}, {11, 12, 13, 14}, {21, 22, 23, 24}}; pri_arr(4, 3, &a[0][0]); }

This method works in all C and C ++ compilers.

But in gcc (C, not C ++), the same example can be written in a more natural way:

 void pri_arr (int ncol, int nlines, int a[][ncol]) { int i, j; for(i = 0; i < nlines; i++) for (j = 0; j < ncol || (puts(""), 0); j++) printf("%d ", a[i][j]); }

Note that the number of elements in an array string must be passed before it is declared in the list of arguments.

 int main() { puts("2 dim"); int a[3][4] = {{1, 2, 3, 4}, {11, 12, 13, 14}, {21, 22, 23, 24}}; pri_arr(4, 3, a); }

When calling, we pass the array itself (and not the address of the first element, as in the first example).

Pavel Parshin Pavel Parshin 8,403 one 15 42 · Answer 3 · 2016-06-05T15:28:23

The first brackets must be left empty, but all of the following must be filled in, indicating the appropriate size. This is necessary so that the compiler can determine the depth of each additional array.

In this way:

 int printmas (int intmas[][4]) {...}

Answer 4 · 2016-06-06T10:13:05

gcc (Ubuntu 4.8.4-2ubuntu1 ~ 14.04) 4.8.4

 #include <stdio.h> #define ROWS 3 #define COLS 4 /*опишите прототип функции. Здесь int (*ar)[COL] указатель на массив из 4 элементов типа int */ void printmas(int (*ar)[COLS] , int rows); int main(void) { /*задайте и проинициализируйте массив*/ int danmassiv [ROWS][COLS]={ {10,9,98,65}, {8,-9,-4,6}, {15,6,78,-8} }; printf("Массив значений имеет вид:\n"); /*вызовите функцию, передав ей имя массива как указатель на 1 элемент(массив из 4 значений типа int) и кол-во элементов в нем */ printmas(danmassiv, ROWS); return 0; } /* определение функции */ void printmas(int ar[][COLS], int rows) { int r; int c; for (r = 0; r < rows; r++) { for (c = 0; c < COLS; c++) /* печать элементов массива,- ar[r][c] указатель на каждый элемент массива */ printf("%d", ar[r][c]); printf("\n"); } }

It is important for you to understand that, in general, arrays and pointers are closely related to each other.
If ar is an array, then the expressions ar [i] and * (ar + i) are equivalent.
As soon as it becomes clear, everything else will fall into place.

Passing a two-dimensional array as a function argument

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