malloc() for Two Dimensional Arrays

Question

I want to initialize a n*n square matrix to later pass it by reference into another function. But it doesn't even compile. I've tried everything, please help.

#include<stdlib.h>
int main()
{
    int i, j, n = 3;
    float **a;
    a = malloc(n * sizeof (float **));//Here I try to create the n pointers to pointer to float for the rows
    for(i = 1;i <= n;i++){
        a[i] = malloc(n * sizeof(float *)); //Here I try to create the n * n pointers to float for the columns
    for(j = 1;j <= n;j++)
        *(*(a + i - 1) + j - 1) = malloc(sizeof(float));  //Here I try to free the space for the elements
    return 0;
}

"It doesn't compile" What error message are you getting, and on what line? — Mooing Duck, Jun 12 '14 at 23:02
You tried *everything*? In Moscow Coding Club, we take candidate's first-born and lock them in basement with rats until algorithms work. *That's* where you see real motivation. — Kerrek SB, Jun 12 '14 at 23:07
|12|error: incompatible types when assigning to type ‘float’ from type ‘void *’| in " *(*(A+i-1)+j-1)=malloc(sizeof(float));" — Ivanz, Jun 12 '14 at 23:08
Dude, instead of `*(A+i)`, how about you just write `A[i]` like everybody else does? — Kerrek SB, Jun 12 '14 at 23:09
@KerrekSB: Ah. Now I understand why Russia is the world champion at ACM ICPC... — nneonneo, Jun 12 '14 at 23:25
in `for(i = 1;i <= n;i++){` you are leaving a `{` opened, you need a closing `}`. — user2485710, Jun 12 '14 at 23:29
yes, it needs "}". It was just an error copying and pasting. In my source code that thing is ok. — Ivanz, Jun 12 '14 at 23:43

score 2 · Answer 1 · answered Jun 12 '14 at 23:48

2

you need to declare it like this :

float **a;
a = malloc(n * sizeof (float *)); // rows
for(i = 0;i < n;i++){
        a[i] = malloc(n * sizeof(float)); // cols
}

answered Jun 12 '14 at 23:48

chouaib

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score 0 · Accepted Answer · answered Jun 12 '14 at 23:43

Your code shows something different to what your comment suggests you are trying to do.

Here I try to create the n * n pointers to float for the columns

Normally a 2-D "array" of float has N x N floats - not N x N pointers to float. You are making a 3-D "array" where the innermost dimension is 1.

Also you are using the wrong type in your sizeof expressions , so you malloc the wrong amount of memory.

Please clarify whether a 2-D "array" is what you want, or if you really do need the extra level of indirection in your code. (I'll update my answer after that).

(NB. I say "array" in quotes as we are not talking about a true multidimensional array; it is in fact an array of pointers which point to the first element of arrays of float (or in the 3-D case, which point to an array of pointers which point to a float).

@Ivanz you mean you want an array of pointers to arrays of float? (a "2D array" would be `float a[n][n];`) — M.M, Jun 13 '14 at 00:06
"Normally a 2-D "array" of float has N x N floats - not N x N pointers to float. You are making a 3-D "array" where the innermost dimension is 1." Bullseye — Ivanz, Jun 13 '14 at 00:30

malloc() for Two Dimensional Arrays

2 Answers2