Dynamically allocate memory for Array of Structs

Question

Here's what I'm trying to do:

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

struct myStruct {
    int myVar;
}

struct myStruct myBigList = null;

void defineMyList(struct myStruct *myArray)
{
     myStruct *myArray = malloc(10 * sizeof(myStruct));

     *myArray[0] = '42';
}

int main()
{
     defineMyList(&myBigList);
}

I'm writing a simple C program to accomplish this. I'm using the GNU99 Xcode 5.0.1 compiler. I've read many examples, and the compiler seems to disagree about where to use the struct tag. Using a struct reference inside the sizeof() command doesn't seem to recognize the struct at all.

n = sizeof( struct myStruct ); and struct myStruct * mBigList = NULL; — Jekyll
– Jekyll, Commented Nov 13, 2013 at 8:25

Ilja Everilä · Accepted Answer · 2017-10-05 10:43:09Z

4

There are a few errors in your code. Make it:

struct myStruct *myBigList = NULL; /* Pointer, and upper-case NULL in C. */

/* Must accept pointer to pointer to change caller's variable. */
void defineMyList(struct myStruct **myArray)
{
     /* Avoid repeating the type name in sizeof. */
     *myArray = malloc(10 * sizeof **myArray);

     /* Access was wrong, must use member name inside structure. */
     (*myArray)[0].myVar = 42;
}

int main()
{
     defineMyList(&myBigList);
     return 0; /* added missing return */
}

Basically you must use the struct keyword unless you typedef it away, and the global variable myBigList had the wrong type.

edited Oct 5, 2017 at 10:43

Ilja Everilä

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answered Nov 13, 2013 at 8:29

unwind

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3 Comments

fkl Over a year ago

+1 for asking to avoid "struct" in sizeof and use the instance name instead.

Mark Löwe Over a year ago

This worked great! Thank you! The ** was the key. Virtual drinks for everyone! Thanks again. All of you cleared up a lot of the confusion.

StarDust Over a year ago

You should free myBigList inside main()

Community · Accepted Answer · 2017-05-23 10:28:15Z

2

This is because struct name is not automatically converted into a type name. In C (not C++) you have to explicitly typedef a type name.

Either use

struct myStruct instance;

when using the type name OR typedef it like this

typedef struct {
    int myVar;
} myStruct;

now myStruct can simply be used as a type name similar to int or any other type.

Note that this is only needed in C. C++ automatically typedefs each struct / class name.

A good convention when extending this to structs containing pointers to the same type is here

edited May 23, 2017 at 10:28

CommunityBot

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answered Nov 13, 2013 at 8:25

fkl

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Comments

sad1e · Accepted Answer · 2013-11-13 08:31:22Z

1

    sizeof(struct myStruct)

or

    typedef struct myStruct myStrut;
    sizeof(myStruct)

answered Nov 13, 2013 at 8:31

sad1e

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Comments

RazvanD · Accepted Answer · 2014-03-03 18:08:47Z

0

In order to work for all 10 elements for that array the line:

myArray[0].myVar = '42';

should be:

(*myArray)[0].myVar = '42';

answered Mar 3, 2014 at 18:08

RazvanD

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2 Comments

Antti Haapala Over a year ago

'42' is not a normal character constant though!

Ilja Everilä Over a year ago

This seems like a comment on the accepted answer, not an answer to the question itself.

Nirav Ambaliya · Accepted Answer · 2016-05-13 04:29:39Z

0

Shouldn't the following statement

myArray[0].myVar = '42';

be this?

(*myArray)[0].myVar = 42;

myvar is an integer.

answered May 13, 2016 at 4:29

Nirav Ambaliya

91 bronze badge

1 Comment

Ilja Everilä Over a year ago

This seems like a comment on the accepted answer, not an answer to the question itself.

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