What is "->" operator called and purpose does it serve?

Question

I'm looking at C-based Solana Smart Contracts and I see -> operator in the code. How do I search for this as I don't even know what to call it (I have tried with no results)?

/**
 * @brief C-based Helloworld BPF program
 */
#include <solana_sdk.h>

uint64_t helloworld(SolParameters *params) {

  if (params->ka_num < 1) {
    sol_log("Greeted account not included in the instruction");
    return ERROR_NOT_ENOUGH_ACCOUNT_KEYS;
  }

  // Get the account to say hello to
  SolAccountInfo *greeted_account = &params->ka[0];

}

thanks, I didn't know "what" to look for. Answer can be found on p203 of The C Programming Language by Kernigan & Ritchie — Shion, Feb 09 '22 at 18:08

score 3 · Accepted Answer · answered Feb 09 '22 at 17:56

3

the -> operator is related to the . operator.

The . operator expects a variable with struct or union type on the left and the name of a member of that struct or union on the right.

The -> does the same thing, except that the left operand is a pointer to a struct or union.

That means that this:

params->ka_num

Is an equivalent and cleaner way of writing this:

(*params).ka_num

answered Feb 09 '22 at 17:56

dbush

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Especially if you need to chain them. `foo->bar->baz` vs. `(*(*foo).bar).baz` – Chris Feb 09 '22 at 18:20

score 1 · Answer 2 · answered Feb 09 '22 at 17:55

1

It must be a pointer dereference operator ->.

answered Feb 09 '22 at 17:55

Ivan Silkin

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What is "->" operator called and purpose does it serve?

2 Answers2