CODEWITHSUNDEEP
c
runners3431
runners3431

Reputation: 1455

how to use pointer notation to access a dynamic struct

I'm confused on how to use pointer notation to access a struct used with malloc. I can use array notation but what's the syntax on pointer? 

  #include "stdio.h"
#include "stdlib.h"

using namespace std;

typedef struct flightType  {

    int altitude;
    int longitude;
    int latitude;
    int heading;

    double airSpeed;

} Flight;

int main() {

    int airbornePlanes;
    Flight *planes;

    printf("How many planes are int he air?");
    scanf("%d", &airbornePlanes);

    planes =  (Flight *) malloc ( airbornePlanes * sizeof(Flight));
    planes[0].altitude = 7;  // plane 0 works

    // how do i accessw ith pointer notation
    //*(planes + 1)->altitude = 8; // not working
    *(planes + 5) -> altitude = 9;
    free(planes);

}

Upvotes: 1

Views: 2159

Answers (3)

johv
johv

Reputation: 4594

Basically x->y is shorthand for (*x).y.

The following are equivalent:

(planes + 5) -> altitude = 9

and

(*(planes + 5)).altitude = 9;

and

planes[5].altitude = 9;

A more concrete example:

Flight my_flight; // Stack-allocated
my_flight.altitude = 9;

Flight* my_flight = (Flight*) malloc(sizeof(Flight)); // Heap-allocated
my_flight->altitude = 9;

Upvotes: 2

shinkou
shinkou

Reputation: 5154

You need an extra parenthesis after dereferencing the pointer.

(*(planes + 5)).altitude = 9;

Upvotes: 0

DrummerB
DrummerB

Reputation: 40211

You don't need the -> notation for this, because you are already dereferencing the pointer using the asterisk. Simply do:

*(places + 5).altitude = 5;

The -> is shorthand for "dereference this struct pointer and access that field", or:

(*myStructPointer).field = value;

is the same as

myStructPointer->field = value;

You can use either notation, but (should) not both at the same time.

Upvotes: 1

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