Suppose model uses all its 30 bytes. Now, make and year want to use also some amount of bytes for their own purposes. But they would overwrite the bytes of model

32 bits

Do you know the difference between a struct and a union?

Can't see much diff actually

That is the problem. In a union, you can set only one of the fields and read back the same field. For ex, if you set "make" in your example, then you can only get "make" out of it. An attempt to read "model" after setting "make" is Undefined Behavior. So at any point of time, only one of the fields will be set. So it is enough to allocate memory large enough to hold the largest type in the union. So your question doesn't arise at all.

So the memory allocated for the entire union is accessible using one and only one of the fields (the one that was lastly set). Got it, thank you. Looks like a redundant data structure.

Why is it redundant?

Why would one need a collection of data the can be accessed exclusively?

To save space when you know only of them is going to be used at any point of time. So the example that you chose "Car" is not a good fit for a union. That should be a struct.

It's like holding a variable this way: char[max_size_needed] various_vars;

This doesnt take alignment into consideration but the idea is correct.

Yeah. BTW, in the event of the union I asked about, the size was of 30 bytes. But what processor alignment was done here? Processors of 32b/64b have to have alignments of 4/8 multiples respectively, no? Whereas 30 bytes is none of those multiplies

No. In your case you only had short and char arrays. The alignment requirement for short is 2 bytes and for char arrays it is byte aligned. So the union will be allocated on an address which is even. It will not be allocated on an odd address.

is there some way to generate automatically a cmake file based on my way of launching app?

but i'm not able to set it in the correct way

that shouldn't be that hard, just make simple cmake, set standard and add the gtk+-3.0 library

In c++ many people mistaken data hiding for data security, what does it mean?

In c++ many people mistaken data hiding for data security, what does it mean?

Write a C program take the pin code from the user and convert in to ascending order using ternary operator(E.g. Input = 382174,Output = 123478)

Write a C program take the pin code from the user and convert in to ascending order using ternary operator(E.g. Input = 382174,Output = 123478)

I dont see a problem

what is the answer

what is the answer

call by reference#include <iostream> using namespace std; int main(void){ int i, j,n,a[100]; i= 0; cout<<"Ingrese un numero: "; cin>>n; while (n > 0){ a[i] = n % 2; n /= 2; ++i; } for (j=0;j<i;++j) cout<<a[j]; cout<<"\n"; system ("pause"); return 0; }

i want to create a function

Function of what? to call a function and pass variable by reference. you simply pass it the address of the variable... so void FunctionName(int *Arg) { *Arg = 10; } And pass it using reference... like so int a = 0; FunctionName(&a);

Function of what? to call a function and pass variable by reference. you simply pass it the address of the variable... so void FunctionName(int *Arg) { *Arg = 10; } And pass it using reference... like so int a = 0; FunctionName(&a);

This is pass by value and not pass by reference

.... i'm a little bit confused now

By default C/C++ passes by value.

Passing a pointer is not pass by reference. C++ has special compound types called references.

Passing a pointer is not pass by reference. C++ has special compound types called references.

I'm sure he is asking about this...

No. This is pass by reference int fun(int& a){ a=5; } int main(){ int a =0; fun(a); }

No. This is pass by reference int fun(int& a){ a=5; } int main(){ int a =0; fun(a); }

sooo, can i do something like this with my code?

Yeah, you're taking reference of the variable when function is called or passing the variable as a reference...

This is what C++ programmers refer to as "pass by reference". Your exple was wrong because that expresses "pass by value" semantics as well. It passes a pointer by value

You could do it either way, both are passing variable by reference.

In C/C++ if variables are passed without and pointers it is pass by value.

You have very little understanding of the standard usage of the terms as you mentioned in a previous post of yours. So please look it up before arguing further

Yep... and I don't want to argue with someone who have a built in hate for me.

I don't have built in hate for you. If correcting you means I have hatred towards you, then by all means assume so. But don't pass of invalid statements are factual here.

Then understand something, you're good but C by default passes variables by value... you don't have to add anything extra.

The question was on C++

Even in C++, it passes all the variables by value by default.

Yes. Even pointers are passed by value. So passing arguments by pointers is not pass by reference in C++