How to free dynamically allocated array?

[Markus]

Hi,

I have read the FAQ 6.16 on how to dynamically allocate a
two-dimensional array.

I do it like this:

double** A = (double **)malloc(m * sizeof(double *));
for(i = 0; i < m; i++)
A = (double *)malloc(n * sizeof(double));

But how do I free it? I currently have only

free(A);

but I guess it is not enough.

The FAQ is not clear about this.

TIA

markus

 

[Krishanu Debnath]

Hi,

I have read the FAQ 6.16 on how to dynamically allocate a
two-dimensional array.

I do it like this:

double** A = (double **)malloc(m * sizeof(double *));
for(i = 0; i < m; i++)
A = (double *)malloc(n * sizeof(double));

But how do I free it? I currently have only

free(A);

but I guess it is not enough.

The FAQ is not clear about this.

for(i = 0; i < m; i++)
free(A);
free(A);

Krishanu

 

[Christopher Benson-Manica]

double** A = (double **)malloc(m * sizeof(double *));
for(i = 0; i < m; i++)
A = (double *)malloc(n * sizeof(double));

Those casts are not necessary (see
http://www.eskimo.com/~scs/C-faq/q7.7.html), and if you've forgotten
to include <stdlib.h>, you will not know it. The practice is better
avoided.

 

[kernelxu]

but I guess it is not enough.

yes, the times you called "malloc()" should be equal to that of calling
"free()".

 

[Flash Gordon]

This is not the preferred form and is error prone. Try
double** A = malloc(m * sizeof *A);

Reasons for this include:
Getting rid of the cast means that all compilers (including C89
compilers are *required* to produce a diagnostic if you forget to
include stdlib.h

By only specifying the type once you only have to get it right once
rather than in the three places you specified type. Think of the risk
of miscounting the *s if you did a 5D array!

It is less to type.

It is less to read

The FAQ at http://www.eskimo.com/~scs/C-faq/q6.16.html is out of date in
this.

Also you need to check if malloc succeeded before you use the space you
hope was allocated. After all, where will the code below write if malloc
failed?

for(i = 0; i < m; i++)
A = (double *)malloc(n * sizeof(double));

A = malloc(n * sizeof *A);

See above.

Correct. Each successful call to malloc needs a corresponding free.

OK, you read the FAQ which is good. You told us that you had read this
section which is even better. It's nice to see evidence that some people
do the right thing.

for(i = 0; i < m; i++)
free(A);
free(A);

This is correct.

 

[Default User]

Hi,

I have read the FAQ 6.16 on how to dynamically allocate a
two-dimensional array.

I do it like this:

double** A = (double **)malloc(m * sizeof(double *));
for(i = 0; i < m; i++)
A = (double *)malloc(n * sizeof(double));

But how do I free it? I currently have only

free(A);

but I guess it is not enough.

The FAQ is not clear about this.

The answer to 7.23 basically covers your question, although that's
about a dynamic structure.

http://www.eskimo.com/~scs/C-faq/q7.23.html


The advice there, one free() for every malloc() is the right one.



Brian

 

[Markus]

This is not the preferred form and is error prone. Try
double** A = malloc(m * sizeof *A);

Reasons for this include:
Getting rid of the cast means that all compilers (including C89
compilers are *required* to produce a diagnostic if you forget to
include stdlib.h

By only specifying the type once you only have to get it right once
rather than in the three places you specified type. Think of the risk
of miscounting the *s if you did a 5D array!

It is less to type.

It is less to read

The FAQ at http://www.eskimo.com/~scs/C-faq/q6.16.html is out of date in
this.

Also you need to check if malloc succeeded before you use the space you
hope was allocated. After all, where will the code below write if malloc
failed?

Thx a lot for your kind advice. Is this proper way to test for malloc
failing?

if(A == NULL)
printf("\nmalloc failed!");

You see I am not good with pointer arithmetic...

markus

 

[Simon Biber]

Thx a lot for your kind advice. Is this proper way to test for malloc
failing?

if(A == NULL)
printf("\nmalloc failed!");

You see I am not good with pointer arithmetic...

Yes, but the \n should go at the end of the line, not the beginning. I
would make A be lowercase, as I reserve all-capitals for macro names.
It's also usual to direct error messages to stderr instead of stdout.

double** a = malloc(m * sizeof *a);

if(a == NULL)
{
fprintf(stderr, "malloc failed!\n");
exit(EXIT_FAILURE);
}

This will exit the program if malloc fails, which may or may not be what
you want your program to do in that case.