c++ - how does the structure dereference operator work? -

I am a Java programmer who is trying to teach me C ++. If I have to ask first question, please let me loose a bit.

I would like to understand how the operator of the structure works. In particular, can someone tell me what the following line of code does in clear words? If (element [i] -> test (arga, argb)) {}

test (arga, argb) is a boolean function Examples of the same class, and element element squared is a vector. Here is the code that immediately encircles the above line, which I'm interested in:

 for  (unsigned i = 0; i  test (arga, argb)) {// some code}} it seems that if the line is testing to see whether the bullion returned or not tested (arga, argb) Element is part of the example given by the class. But when I try to highlight the basic values ​​of the above cout lines of elements [i] or test (arga, argb), the compiler throws errors until I comment out those lines. In Java, I would be able to fiddle around it until I could not meet the values ​​of each other, and then I would understand the code line. But I do not know what this line of code does in C ++. Does anyone give me a clear explanation, which can be linked to some references or more than two supported?   
 
 
  element [i] - & gt; Exam (arga, argb)   
 
 If we break the statement,  from left to right , we will end up with:  < Ol> 
 
 Use the  i  element in an array (or  array-like ) element  element   < / Li> 
 
 The element's access ( element [i] ) is an indicator for an object   
  member-function  designated  test  on the  element [i]  and pass it two arguments;  RGA  and  argb    
 
 If we ignore the fact that you have written  std :: cout & Gt; & Gt;  std :: cout & lt; & Lt; Instead of  (the next is the correct form), we conclude the two of the errors mentioned by you:  
 
 
 About your compiler  std :: Cout & lt; & Lt; Element [i]  because  element [i]  and  std :: ostream & amp; There is no appropriate overload (which is the underlying type of  std :: cout ) to control the type of .   
 
 Your compiler is  std :: cout  because  test  has no function in the scope of the name, which is  arga, argv  test code , in your snippet, an  member-function  that comes under a single unit, calling it by itself Deserving Is not.