Re: In map iterator is there a difference between (*iter).second and iter->second?

From: (Pascal J. Bourguignon)
Wed, 20 Aug 2008 11:42:00 +0200
"Mike Wahler" <> writes:

"puzzlecracker" <> wrote in message

I see that a lot of former in the code, and wonder if there is a
technical reason for that

You're asking about language syntax, this is
not specific to map or iterators.

Given a pointer to a class or struct:

struct T
    int member;

and a pointer to one of these structs:
T obj;
T *p(&obj);

The two expressions:




have exactly the same meaning.

The '->' form is 'shorthand' for the (*). form.

Yes, but given a different class, they may be different:

#include <iostream>

class A {
public: int x;

class P {
    A a1;
    A a2;

    A& operator*(){ return(a1); }
    A* operator->(){ return(&a2); }

using namespace std;

int main(void){
    P p;
    cout<<"p->x = "<<p->x<<" ; (*p).x = "<<(*p).x<<endl;

-*- mode: compilation; default-directory: "~/src/tests-c++/" -*-
Compilation started at Wed Aug 20 11:40:06

SRC="/home/pjb/src/tests-c++/memb.c++" ; EXE="memb" ; g++ -g3 -ggdb3 -o ${EXE} ${SRC} && ./${EXE} && echo status = $?
p->x = 2 ; (*p).x = 1
status = 0

Compilation finished at Wed Aug 20 11:40:07

Why someone used the latter over the former I
don't know. Perhaps a 'style' issue, or possibly
that's what a code generator created.

Or perhaps they mean different things.

__Pascal Bourguignon__

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