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

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.

CodePudding user response：

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;

CodePudding user response：

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'
}