help with declaring array in which the user specifies size

**ibanezrgking** · 02-10-2002

here's what i want to do:

program asks user how big the array is gonna be.

cout << "how big you want this array to be?" << endl;

cin >> array_size;

user types: 10(or however large he wants the array to be)[enter]

program makes an array of size n:

int array[array_size];

BUT! alas......"error C2057: expected constant expression"

the error occurs on this line: int array[array_size];

i know theres a way to do this. and i also know this isnt the way....but its been a long time since i made progs...so someone please refresh my atrophied memory?

**Salem** · 02-10-2002

cin >> array_size;
int *array = new int[array_size];

Then you do stuff like
for ( int i = 0 ; i < array_size ; i++ ) array[i] = 0;

When you're done, tidy up with
delete [] array;

**ibanezrgking** · 02-10-2002

thanks, salem

but i dont know what exactly new or delete does....i feel like im gonna have trouble implementing those things without knowing what they do...and i saw that pointer * whats it for...its been too long...sorry

can you help elaborate on what that code does exactly?

appreciate it lots.

edit: YAY! i just found a tutorial on dynamic memory! i think its gonna answer some questions.

**samGwilliam** · 02-10-2002

Well, as an array is essentially a pointer to a contiguous memory block, we can delcare an int pointer instead of an array. The new operator will return the address of a block of memory the size of what you wish to allocate, therefore your pointer becomes an array of whatever size you wish! As this was allocated dynamically, the program has no record of it so it's up to you to free up the memory with the delete operator, or else you'll get leaky code.

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