I cannot access the elements of an array of structure with ->

Question

I'm having a little problem with a function which receives as a parameter an array of structures, the problem occurs when trying to access array elements with operator ->

#include <stdio.h>

typedef struct{
    int order;
}record;

void entry(record*reg, size_t num_regs);

int main(void){
    record reg[10];
    entry(reg, sizeof reg / sizeof reg[0]);

    return 0;
}

void entry(record*reg, size_t num_regs){
    size_t i;

    for (i = 0; i < num_regs; i++){
        reg[i]->order = i;
        printf("\n order = %d", reg[i]->order);
    }
}

throws this error if you try to compile

*error #2140: Type error in argument 1 to 'ingreso'; expected 'registro * *' but found 'registro *'.*

because it throws this error and how to fix it?

Answer 1

When you use [ ] for pointer you already have a data, so use . instead of ->

reg[i].order = i;

and the same for printf argument.

Answer 2

I tried your code like this and whole of the array printed exactly true:

for (i = 0; i < num_regs; i++){
    reg->order = i;
    printf("\n order = %d", reg->order);
}

Answer 3

You need to understand the difference between the -> and . operators when accessing data inside of a struct .

• sa is just for when s is a struct and a is a member of s
• sp->a is really just a shorthand for (*sp).a This is used when sp is a pointer-to- struct , and we want to dereference that pointer and access the struct 's data all in one step.

As VolAnd said , you are using reg[i]->order but you should really be using reg[i].order .

---

If it's still unclear...

With your entry() function, you are passing in an array of 10 record structs called reg . You do this by passing a pointer to the array's base address, so the function accepts a pointer of type record* .

The elements of the array are structs , not struct pointers , so you access them with reg[i].order , not reg[i]->order .

Answer 4

Here you are passing address of array and storing / collecting is in pointer. And as the pointer is pointing to array, if you use -> operator to access the structure element it will treat array as pointer and throws error as "error: invalid type argument of â->â (have ârecordâ)".

So you have to use (.) operator instade of (->) operator.

for (i = 0; i < num_regs; i++)
{
    reg[i].order = i;
    printf("\n order = %d", reg[i].order);
}