Tried to do 3 dimensional matrix using malloc

Question

I want to collect a 3-dimensional matrix using malloc. There is also the matrix printing part but I did not include that.

#include <stdio.h>
#include <stdlib.h>
int main()
{
   int i,j,k,m,n,o;
   float ***A;
   printf("Input dimension for matrix(m,n,o)\n");
   printf("Enter m : ");
   scanf("%d",&m);
   printf("Enter n : ");
   scanf("%d",&n);
   printf("Enter o : ");
   scanf("%d",&o);
   A = (float***)malloc(o*sizeof(float**));
   for(i=0;i<o;i++){
           A[i] = (float**)malloc(m*sizeof(float*));
           for(j=0;j<m;j++){
               A[i][j] = (float*)malloc(n*sizeof(float));
           }
   }
   for(i=0;i<o;i++){
       for(j=0;j<m;j++){
           for(k<0;k<n;k++){
               printf("Input number for (%d,%d,%d) : ",j+1,k+1,i+1);
               scanf("%f",&A[i][j][k]);
           }
       }
   }
   return 0;
}

Program seems to be skipping to the end after I finish input m,n,o, so I cannot enter any value of matrix. I have to use malloc because it was required for the task.

EDIT: everything that is wrong. m is for row,n is for column, and o is for a number of mXn matrix.

you seem to be using different numbers for malloc and for your `for` loops. E.g. if dimensions are 111,222,333 then you allocate 111 float**s and then put 222 pointers into that space. — user253751, Sep 08 '21 at 13:18
You've confused the issue by editing the code. Which is the *actual* code you have used? Please copy/paste the exact code you used. — Weather Vane, Sep 08 '21 at 13:21
@WeatherVane I edit of what you ask in this and actual code. But it didn't fix. — Secret Number, Sep 08 '21 at 13:24
Watch out for compiler warnings. `scanf("%f",A[i][j][k])` is wrong, it should be `&A[i][j][k]` — Weather Vane, Sep 08 '21 at 13:27
Secret Number, Are you required to use the [unnecessary cast](https://stackoverflow.com/a/605856/2410359) with `malloc()`? — chux - Reinstate Monica, Sep 08 '21 at 14:20
You could optionally just do a `malloc(m * n * o * sizeof(float))` and just do some simple arithmetic to find the correct index. I prefer this for structures like these, because it will be a single `malloc` and a single `free`, no nasty loops. — Cheatah, Sep 08 '21 at 15:02

score 3 · Answer 1 · edited Sep 08 '21 at 14:04

3

Use dynamic Variable-Length Array:

float (*A)[n][m] = calloc(o, sizeof *A);

And free it at the end with

free(A);

More one must passed an address of the scanf-ed object, not it's value. It should be:

scanf("%f", &A[i][j][k])

edited Sep 08 '21 at 14:04

Robert Harvey

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answered Sep 08 '21 at 13:31

tstanisl

13,520
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Sorry. But float A*** cannot be remove from the code. So i can't using the dynamic VLA. – Secret Number Sep 08 '21 at 13:36
@SecretNumber This is away to allocate for a 3D array. `float A***` approach is _not_ a 3D array, but an array of pointers to an array of pointers to an array of `float`. – chux - Reinstate Monica Sep 08 '21 at 13:45
@chux-ReinstateMonica My homework task is fixed for `float A***`. So I have to used that method instead. – Secret Number Sep 08 '21 at 13:58
@SecretNumber OK, so you do not want a true 3D array as titled, but a [jagged array](https://en.wikipedia.org/wiki/Jagged_array). Sorry if your course is not using crisp terms. – chux - Reinstate Monica Sep 08 '21 at 14:00

nilTheDev · Accepted Answer · 2021-09-08T14:12:21.123

I think all you did is messed up in the for loop while taking the inputs. As per the context, the 3D matrix, I debugged the code and came up with a version that makes sense.

I will go on and point out the mistakes you made one by one.

First, scanf takes the memory address of the variable not the dereferenced version of it.

So, change it from this, scanf("%f",A[i][j][k]); to this scanf("%f",&(A[i][j][k]));. This one would also work scanf("%f",A[i][j] + k);.

In this line, for(k<0;k<n;k++), k is not even initialized. So you are essentially operating on a garbage value. As my mind was already baffled by deciphering the c pointer hell, I couldn't even process the end result. Code that would make more sense is this for(k=0;k<n;k++).

After that, you switched the control logic of the nested for loops.

Replace the variable m in this code for(j=0;j<m;j++) with n. for(j=0;j<n;j++)

Do vice-versa in the next loop. Modify this for(k=0;k<n;k++) to this for(k=0;k<m;k++).

The final code would look like this,

#include <stdio.h>
#include <stdlib.h>
int main()
{
   int i,j,k,m,n,o;
   float ***A;
   printf("Input dimension for matrix(m,n,o)\n");
   printf("Enter m : ");
   scanf("%d",&m);
   printf("Enter n : ");
   scanf("%d",&n);
   printf("Enter o : ");
   scanf("%d",&o);
   A = (float***)malloc(o*sizeof(float**));
   for(i=0;i<o;i++){
           A[i] = (float**)malloc(n*sizeof(float*));
           for(j=0;j<n;j++){
               A[i][j] = (float*)malloc(m*sizeof(float));
           }
   }
   for(i=0;i<o;i++){
       for(j=0;j<n;j++){
           for(k=0;k<m;k++){
               printf("Input number for (%d,%d,%d) : ", i + 1, j + 1, k + 1);
               scanf("%f",&(A[i][j][k]));
           }
       }
   }
   
   
   return 0;
}

It would now work as a 3D matrix.

To understand the code I had to draw a messy diagram. I am attaching that. It may be of some help.

As you have already omitted some code I hope you have not forgotten to use free to prevent memory leaks.

Oh yeah. I didn't see the k<0 one. Thank you so much – Secret Number Sep 08 '21 at 14:15 — Secret Number, Sep 08 '21 at 14:15

Luca Polito · Answer 3 · 2021-09-08T14:07:01.417

0

This should work:

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

int main(void) {
    float ***A;

    // n1, n2 and n3 will contain the three dimensions (i.e.: A[n1][n2][n3])
    size_t n1, n2, n3;

    printf("Insert first dimension: ");
    if (scanf("%zu", &n1) != 1) {
        printf("ERROR: cannot read number\n");
        return 1;
    }

    printf("Insert second dimension: ");
    if (scanf("%zu", &n2) != 1) {
        printf("ERROR: cannot read number\n");
        return 1;
    }

    printf("Insert third dimension: ");
    if (scanf("%zu", &n3) != 1) {
        printf("ERROR: cannot read number\n");
        return 1;
    }

    // firstly, here we alloc space for first dimension
    A = malloc(n1 * sizeof(float**));

    for (size_t i = 0; i < n1; i += 1) {
        // now we alloc space for second dimension
        A[i] = malloc(n2 * sizeof(float*));
        for (size_t j = 0; j < n2; j += 1) {
            // here we alloc space for third dimension
            A[i][j] = malloc(n3 * sizeof(float));
        }
    }

    // let's ask the user to input all elements into the array
    for (size_t i = 0; i < n1; i += 1) {
        for (size_t j = 0; j < n2; j += 1) {
            for (size_t k = 0; k < n3; k += 1) {
                printf("Insert value for A[%zu][%zu][%zu]: ", i, j, k);
                if (scanf("%f", &(A[i][j][k])) != 1) {
                    printf("ERROR: cannot read number\n");
                    return 1;
                }
            }
        }
    }

    // now let's print all the numbers that the user has input previously
    for (size_t i = 0; i < n1; i += 1) {
        for (size_t j = 0; j < n2; j += 1) {
            for (size_t k = 0; k < n3; k += 1) {
                printf("A[%zu][%zu][%zu] = %f\n", i, j, k, A[i][j][k]);
            }
        }
    }

    return 0;
}

I've not compiled it, but it should be correct. Take a close look at how allocation has been made.

NOTE: of course, total correctness (e.g.: use of free() and avoidance of scanf()) has been avoided on purpose.

P.S.: as other users already mentioned, take care to enable warnings for your compiler (e.g.: -Wall -Wextra on GCC), because it will simplify debugging very much for you.

edited Sep 08 '21 at 14:07

answered Sep 08 '21 at 13:43

Luca Polito

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Rather than allocate to the size of type, allocating to the size of the referenced object is less error prone and easier to maintain. `arr = malloc(n1 * sizeof *arr);` – chux - Reinstate Monica Sep 08 '21 at 13:47
@chux-ReinstateMonica Yes, yours is one good alternative approach. – Luca Polito Sep 08 '21 at 13:50
@LucaPolito Sorry. But as I told above. I can't change `float A***` to anything else. – Secret Number Sep 08 '21 at 13:59
@SecretNumber This answer uses `float***`. How does that not fulfill your goals? – chux - Reinstate Monica Sep 08 '21 at 14:03
@SecretNumber Look better. I haven't changed your `float ***A`, I've just renamed it to `arr`. You can replace `arr` with `A` if you want. – Luca Polito Sep 08 '21 at 14:04
I can't add anything after `float ***A` too. This line was fix and cannot edit or add anything. – Secret Number Sep 08 '21 at 14:07
@SecretNumber I've renamed `arr` to `A`, the code remains the same. – Luca Polito Sep 08 '21 at 14:07
`float*** arr = malloc(n1 * sizeof(float**));` I can't do this. But I can if you remove `float***` – Secret Number Sep 08 '21 at 14:10
@SecretNumber I've edited the code already. – Luca Polito Sep 08 '21 at 14:10
@SecretNumber P.S.: `float ***A = ...;` is just a shortcut for `float ***A; A = ...;` – Luca Polito Sep 08 '21 at 14:11

Tried to do 3 dimensional matrix using malloc

3 Answers3