dynamically allocating an array for a matrix

Question

So i'm trying to allocat an array to hold a matrix of floats. the values passed in are ints (rows and cols) and the function is a pointer. so this is my function definition:

float *matrix(int rows,int cols)
{
    int i=0;
    float *m=NULL;
    m=(float *)malloc(rows*sizeof(int));
    for (i=0;i<rows;i++)
    {
        m[i]=(float*)malloc(cols*sizeof(int));
    }   
}

i have a feeling that this is wrong. i also get an error when i try to run. where exactly is the problem here? should the int be a float instead?

edit****

float *matrix(int rows,int cols)
{
    int i=0;
    float **m=NULL;
    m=(float *)malloc(rows*sizeof(float));
    for (i=0;i<rows;i++)
    {
        m[i]=(float *)malloc(cols*sizeof(float));
    }
}

alright if malloc doesn't run properly and fails i wan't to return NULL. so it should be this code here, right?

if(m[i]==NULL)
{
    return NULL;
}

A float* isn't an array of float*s.

tkausl
– tkausl

2016-07-02 02:52:51 +00:00
Commented Jul 2, 2016 at 2:52 — tkausl
– tkausl, Commented Jul 2, 2016 at 2:52

Community · Accepted Answer · 2017-05-23 11:44:16Z

1

    int i=0;
    float **m; // You are looking for a pointer to pointer to float
    m=malloc(rows*sizeof(float*)); //Step1
    for (i=0;i<rows;i++)
    {
        m[i]=malloc(cols*sizeof(float)); //Step2
    }

Notes

You need not the cast the output of malloc for the reason mentioned in [ this ] answer.
In step1, you allocate memory for rows float*s
For each of the float* in step1, we allocate memory to store cols floats in step2

edited May 23, 2017 at 11:44

CommunityBot

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answered Jul 2, 2016 at 2:58

sjsam

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