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

Question

I am a java programmer trying to teach myself c++. Please cut me a little slack if I ask simple questions at first.
I would like to understand how the structure dereference operator works. Specifically, can anyone tell me what the following line of code does in explicit terms?
if (elements[i]->test(arga, argb)) {}
test(arga,argb) is a Boolean function in the same class, and elements is a vector of instances of the element class. Here is the code that immediately surrounds the line above, about which I am interested:
for (unsigned i = 0; i < elements.size(); ++i) {
T arga = INFINITY, argb = INFINITY;
//using namespace std;
//std::cout >> elements[i] >> std::endl;
//std::cout >> test(arga, argb) >> std::endl;
if (elements[i]->test(arga, argb)) {
//some code
}
}
It seems that the if line is testing to see whether or not the boolean returned by test(arga,argb) is part of the given instance of the elements class. But when I try to expose the underlying values of elements[i] or test(arga,argb) with the cout lines above, the compiler throws errors until I comment those lines out. In java, I would be able to fiddle around with this until I found values of each that correspond with each other, and then I would understand the line of code. But I do not know how to figure out what this line of code does in C++. Can anyone give me a clear explanation, preferably supported by a link or two to some references online?

Answer 1

elements[i]->test (arga, argb)
If we break down the statement, reading from left-to-right, we will end up with the below:
access the ith element in an array (or array-like) entity named elements
the element accessed (elements[i]) is a pointer to an object
call the member-function named test on elements[i] and pass it two arguments; arga and argb
if we disregard the fact that you wrote std::cout >> instead of std::cout << (the latter is the correct form), we end up with two reasons for your described errors:
your compiler complains about std::cout << element[i] because no suitable overload is found to handle an entity of the type of element[i] and an std::ostream& (which is the underlying type of std::cout).
your compiler complains about std::cout << test (arga, argb) because there is no function in scope named test that takes two arguments corresponding to arga, argv. test, in your snippet, is a member-function that belongs to an entity, it's not callable by its own.

Answer 2

Welcome to C++.
First, the syntax for output is:
cout<<
instead of
cout>>
You are right in guessing that test is a function that returns boolean.Here elements[i] is a pointer pointing to a struct element which has this test function.
To learn C++, you can use these articles that I wrote.Good luck!

Answer 3

Since numerous respondents told me that I need to provide the code before they can answer, I looked deeper in the code, and re-wrote something which tells me that the line:
if (elements[i]->test(arga, argb)) {}
is a test to see whether or not the boolean member function of elements[i] is true.
The c++ program that I wrote to identify the meaning of -> in this context is:
#include "stdafx.h"
#include <vector>
#include <string>
#include <iostream>
template<typename T>
class Bone
{
public:
std::string myName;
int mySize;
Bone(const std::string &name, const int &size) : myName(name), mySize(size)
{}
bool isBigger(const int &testSize) const
{
if (testSize > mySize) return false;
else return true;
}
};
int main(int argc, char **argv)
{
std::vector<Bone<float> *> bones;
// name, location, size
bones.push_back(new Bone<float>("femur", 10));
bones.push_back(new Bone<float>("ulna", 4));
bones.push_back(new Bone<float>("maxilla", 3));
int testSize = 6;
// test each bone to see if it is bigger than testSize
for (unsigned i = 0; i < bones.size(); ++i) {
if (bones[i]->isBigger(testSize)) {
std::cout << bones[i]->myName; std::cout << " is bigger than testSize! " << std::endl;
}
}
while (!bones.empty()) {
Bone<float> *thisBone = bones.back();
bones.pop_back();
delete thisBone;
}
return 0;
}
Thank you to everyone who led me to figure this out.