how to do dynamic allocation in boundless array - c

Question

Well I am wanting to change the way my structures are written, currently I use array and I need to limit its use, but I wanted a way to create a dynamic array that is the size of the reading done, without always having to edit the array value.
Current Code:
struct sr_flag {
int value_flag;
};
struct er_time {
int value_time;
};
struct se_option {
struct sr_flag flag[50];
struct er_time time[50];
};
struct read_funcs
struct se_option *option;
void (*option_func) (void);
...
}
struct read_funcs func_;
struct read_funcs *func;
int sr_flags(int i, int fg, int val) {
if(i < 0)
return 0;
return func->option[i].flag[fg].value_flag = val;
}
void option_func(void) {
struct se_option fnc;
fnc.option = malloc(500 * sizeof(*(fnc.option)));
}
void read_fnc() {
func = &func_;
func->option = NULL;
func->option_func = option_func;
}
I look for a way to remove the array amount [50] instead each time the sr_flags function is executed the limit is raised
Example: sr_flags function executed 1x array would be [1] if executed 2x would be [2]
I also think about doing the same with the option_func function
I tried using the following more unsuccessfully
struct se_option {
struct sr_flag *flag;
struct er_time time[50];
};
int sr_flags(int i, int fg, int val) {
if(i < 0)
return 0;
func->option[i].flag = malloc(1 * sizeof(*(func->option[i].flag)));
return func->option[i].flag[fg].value_flag = val;
}
int main () {
for(int i < 0; i < 10; i++)
sr_flags(i, 1, 30);
return 0;
}

Answer 1

I'm not 100% certain on what it is you want but I think you just want to call realloc and increase the size by the amount you provide. And that's very easy to do, as for the values you want with the arrays I'm not sure so I just used a placeholder value.
#include <stdio.h>
#include <stdlib.h>
struct sr_flag {
int value_flag;
};
struct er_time {
int value_time;
};
struct se_option {
struct sr_flag* flag;
struct er_time* time;
};
void allocateflags(struct se_option* options, int size, int val){
options->flag = realloc(options->flag, size*sizeof(struct sr_flag));
struct sr_flag* flag = options->flag+size-1;
flag->value_flag = val;
}
void allocatetime(struct se_option* options,int size, int val){
options->time = realloc(options->time, size*sizeof(struct er_time));
struct er_time* time = options->time+size-1;
time->value_time = val;
}
void displayflagvalues(struct se_option* options,int size){
for(int index = 0; index < size ; ++index){
printf("flag: %i\n",options->flag[index].value_flag);
}
}
void displaytimevalues(struct se_option* options, int size){
for(int index = 0; index < size ; ++index){
printf("time: %i\n",options->time[index].value_time);
}
}
int main(){
struct se_option options = {0};
for(int index = 0; index < 10; ++index){
allocateflags(&options, index,index);
allocatetime(&options, index,index);
}
displayflagvalues(&options, 10);
displaytimevalues(&options,10);
return 0;
}
The code creates an se_option structure wheren sr_flag and er_time pointers are null. Then there's two functions one allocateflags and the other allocatetime, both of which call realloc with the size you provide. When you call realloc, all previous memory is copied over to the new array. Also free is called automatically by realloc.
This step
struct sr_flag* flag = options->flag+size-1;
flag->value_flag = val;
struct er_time* time = options->time+size-1;
time->value_time = val;
Is slightly redundant but it was just to show the newest array can hold the value. If you understand pointer arithmetic, all its doing is incrementing the pointer to the last position then subtracting 1 struct size and setting that value. Basically setting the value of the final array in the pointer.