Question

I don’t understand why I’m getting this error. I used the same code previously and it gave me no problem, but now for some reason im getting an error.

            outputf3 << "The sorted list is as follows: \n\n";
            sorting = new float [function.sort()]; // Here is where the error is.
            for (i=0; i < length ; i ++)            
            {
            outputf2 << "\n" << sorting[i] ;
            }

For some reason it’s saying
error: expression in new-declarator must have integral or enumeration type

But i used the same thing here in a previous part of the program.

           inputs = new float[length];
           inputf.clear();
           inputf.seekg(0, std::ios_base::beg);

I have no idea why it’s doing this at all.
What I’m doing is reading random numbers in from a file and then sorting them in another part of the function and writeing them out to the output file.

Answer 1

Why do you think:

inputs = new float[length];

is the same as

sorting = new float [function.sort()];

?

Unless function is a class instance and sort returns an integral type, it’s certain it won’t work.

The statement:

Type* t = new Type[x];

dynamically allocates x objects of type Type. It’s fairly obvious why x has to be of integral type.

EDIT: As per your comment:

The size to be allocated needs to be of integral type, but you return a float. But even then, it wouldn’t make sense. You need to return the size of inputs.