Problem with setting the array of pointers in the struct using malloc

Question

I want to allocate a memory to an array of pointers in struct, but I receive the following error:

expression must be a modifiable lvalue

Here's struct code:

typedef struct {
    int id;
    char *entity[];
}entity;

Here's memory allocation in main function:

entity s;
s.entity= malloc(30 * sizeof(char *));

IDE underlines s.entity and pops the error I mentioned.

Please, help me out to solve this issue.

an array is not modifiable. What you want instead is char** set — Raildex
– Raildex, Commented Nov 14, 2021 at 13:07
The way you wrote this, you should basically use: entity *s = malloc(sizeof *s + 30 * sizeof *s->set); — Cheatah
– Cheatah, Commented Nov 14, 2021 at 13:10

0___________ · Accepted Answer · 2021-11-14 13:27:03Z

4

Your structure does not have a member called entity, only id and set.

You apparently want to allocate the whole structure. This type of struct member called flexible array member is useful if you want to allocate the whole structure in one malloc.

entity *s;
s = malloc(sizeof(*s) + 30 * sizeof(s -> set[0]));

This kind of struct members are very useful as you can realloc or free them in a single call.

Increase the size of the set array to 50

entity *tmp = realloc(s, sizeof(*s) + 50 * sizeof(s -> set[0]));
if(tmp) s = tmp;

edited Nov 14, 2021 at 13:27

answered Nov 14, 2021 at 13:14

0___________

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3 Comments

Cheatah Over a year ago

I am not one of the downvoters, but your first suggestion didn't work because set is a flexible array member. I guess you realised, since you edited your answer.

Eric Postpischil Over a year ago

@Cheatah: What do you see wrong in the first version of the answer? It has malloc(sizeof(*s) + 30 * sizeof(s -> set[0]));, which is correct for allocating space for the structure plus 30 elements in the array member.

0___________ Over a year ago

@EricPostpischil the very first one was wrong :)

vmp · Accepted Answer · 2021-11-14 13:20:35Z

1

Thats how you would allocate the pointers:

typedef struct {
    int id;
    char **set;
}entity;

int how_many_pointers = 30;
entity s;
s.set= malloc(how_many_pointers * sizeof(char *));

And for each pointer you would have to allocate the space for the corresponding string:

int i, string_size;

for(i = 0; i < how_many_pointers; i++)
{
    printf("How many chars should have string number %d ?", i + 1);
    scanf("%d", &string_size);
    s.set[i] = malloc((string_size + 1) * sizeof(char)); // string + 1 due to space for '\0'
}

answered Nov 14, 2021 at 13:20

vmp

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