C / C ++指针解除引用

Question

If I have the following function...

void function(double *array)
{
    double a = array[3]; // Method 1
    double b = *(array + 3); // Method 2
}

Assume the array has 5 elements (I do know the length of the array ahead of time). The code compiles fine and runs ok. 'a' and 'b' do contain the expected value.

In what instances would I use method 2 as opposed to method 1?

Answer 1

E1[E2] is equivalent in C to (*((E1) + (E2))) by definition of [] operator.

Prefer the first notation as it is shorter and more readable.

Answer 2

both do the same. that's like asking which method is better for accessing struct member allocated on heap :

#include <stdio.h>
#include <stdlib.h>

struct s { int a;};
int main(void)
{
    struct s* x = malloc(sizeof(struct s));
    x->a = 2;  //method 1 OR
    printf("%d\n",x->a);
    (*x).a = 3;  //method 2
    printf("%d\n",(*x).a);
    free(x);
    return 0;
}

so use [] and -> operators. it's good to know how they really work (dereferencing a pointer and adding offset to get to right address).