E-> ?

This is a discussion on E-> ? within the C++ Programming forums, part of the General Programming Boards category; Hello, I'm trying to understand what's going on in a certain program I need to use (not actually sure if ...

  1. #1
    c_h
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    E-> ?

    Hello,
    I'm trying to understand what's going on in a certain program I need to use (not actually sure if it's written in C or C++). I haven't worked with C/C++ before, but I picked up a manual and can follow most of what's going. What I'm confused about:

    fprintf(E->filed[13], "%6d/n", E->lmesh.nno);

    so the only part of the above statement that I don't understand is this E-> , could anyone give me an idea as to what's going on here or at least tell me what this is called so I can look it up in my manual?

    Thanks in advance!

  2. #2
    C++まいる!Cをこわせ! Elysia's Avatar
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    E is the variable, which is a pointer.
    And the "->" syntax is the a shortcut for "(*var)."
    So it's the same as writing (*E).filed[13].
    Quote Originally Posted by Adak View Post
    io.h certainly IS included in some modern compilers. It is no longer part of the standard for C, but it is nevertheless, included in the very latest Pelles C versions.
    Quote Originally Posted by Salem View Post
    You mean it's included as a crutch to help ancient programmers limp along without them having to relearn too much.

    Outside of your DOS world, your header file is meaningless.
    For information on how to enable C++11 on your compiler, look here.
    よく聞くがいい!私は天才だからね! ^_^

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    c_h
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    thanks!

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