C++ Initializing Struct Without Pointer

Question

I have a struct in C++

typedef struct DIFColor
{
    uint8_t r;
    uint8_t g;
    uint8_t b;


} DIFColor;

And am trying to initialize it like this:

DIFColor d = new DIFColor();

d.r = r;
d.g = g;
d.b = b;

But the compiler complains about not being able to convert a pointer to a regular variable. What pointer? From this question I thought you could simply create a new struct without having to deal with pointers and dynamic memory allocation with new. Does it have something to do with the fact that I am initializing the struct like one would in C, instead of using a constructor?

error: conversion from 'DIFColor*' to non-scalar type 'DIFColor' requested

Answer 1

Try:

DIFColor d = { r, g, b };

Answer 2

The operator new will return a pointer to your struct. If you want to declare the object on the heap

DIFColor* d = new DIFColor;
d->r = r;
d->g = g;
d->b = b;

Or to declare it on the stack don't use new.

DIFColor d;
d.r = r;
d.g = g;
d.b = b;

Answer 3

You should not use dynamic allocation without reason, and there is no reason in your code, so remove new:

DIFColor d = new DIFColor();

replace with

DIFColor d;

Answer 4

The new keyword in

DIFColor d = new DIFColor();

creates a pointer. You should either write

DIFColor d = {r,g,b}; // allocate on stack

or

DIFColor * d = new DIFColor(); // make a pointer

Answer 5

new operator returns a pointer.

So, correct syntax would be:

DIFColor* d = new DIFColor();