Allocating size for a variable in c++

You mean you don't know the actual size ahead of time? Both malloc() and new support the use of variables as input.

//Example...
int *a = (int*) malloc(x*sizeof(int));
//Where x can be any positive integer

edit:
by the way... if you don't know the size because it's going to keep changing, then you can use something with a dynamic size like a list instead of a fixed size array.

One of the thing I will never understand is why who is learning C++ is always tempted (or directed to) learn as it is C.

C++ support classes and abstract types and has a standard library whose specification is part of the language and that should be studied together with the language.

Give a look to std::vector[^].

All you need is a

std::vector<int> a;

and then, every time you have a value to insert in it, just call

a.push_back(your_value);

You can then know the size with a.size() and access the values as usual as a[x].

Hi,


You can't initialize without using the size.
You need to init it with a size then expand you buffer on your need.
You need to use realloc function to expand your memory buffer :

MSDN realloc[^]


[EDIT]
you should have a kind of linked list but with different data type ... look at this example it could inspire you :
linked list[^]


Best regards.
EL GAABEB.

I assume you want an array of ints?
Assumming an int is 4 bytes, you just allocated 400 bytes to a array of 100 ints.

Now beyond the obvious, the answer is, "it depends".

Sometimes you have to loop through your data and calculate how much you need.
Ex, concatenating two strings into one, you need to get the length of both strings, allocate a buffer large enough to hold both, then copy both in appropriately.

Some APIs have functions you call once to get the size, allocate the amount of data you were given, then call the function again to get the data out.

if you want to allocate on the byte level, allocate char instead.

Unless you get more specific, I am not sure how else to help you.