size of dynamic array

[kenneth6]

int* x=new int[10];
cout << (sizeof(x)/sizeof(int)) << endl;
x[0]=10;
x[1]=20;
x[2]=30;
x[3]=40;
cout << x[0] << endl;
cout << x[1] << endl;
cout << x[2] << endl;
cout << x[3] << endl;
cout << x[4] << endl;
cout << (sizeof(x)/sizeof(int)) << endl;

Why can't i get the number of elements in this array?

[Laharl]

sizeof, when used on a pointer, returns the size of the pointer. The sizeof of a pointer is the number of bytes in one 'word' of the computer, 4 for a 32bit machine. An int is also 4 bytes on a 32bit machine, so that will print out 1 everytime on nearly every computer.

[Ganon11]

The convenient, simple, and most often used method for figuring out the size of an array is...not to figure it out. Make it a separate variable, or if it's a function, require an additional argument indicating the size. Especially for the function, if someone gives you the wrong size, well, that's their own fault.

[kenneth6]

cout << x[4] << endl;      // within the max boundary of array, but not initialized
cout << x[10] << endl;     // beyond the max boundary of array
cout << x[11] << endl;
cout << x[12] << endl;

if i print the above three lines in the above situations, what numbers will be printed out? are the numbers meaningful?

[Ganon11]

No one knows. When you create an array, all it is is a block of memory. So when you say:

[code=cpp]int myArray = new int[10];[/code]

somewhere in memory, there is a space 10 * sizeof(int) long (probably 40 bytes):

 [memory][myArray[0]][myArray[1]]...[myArray[9]][more memory]...

So if you use myArray[10] or anything else past the bounds, you are looking into memory which may or may not be allocated for your program. You have no idea what you might be printing out. You'll be lucky not to get a segmentation fault for accessing illegal memory locations.

[Laharl]

You'll get garbage, whatever happened to reside in memory at those locations. If you're lucky, that is. If you're not, it's possible to get a segfault from that.

[kenneth6]

If you use myArray[10] or anything else past the bounds, you are looking into memory which may or may not be allocated for your program. You have no idea what you might be printing out. You'll be lucky not to get a segmentation fault for accessing illegal memory locations.

Thanks! I don't know about the hardware memory. May I know if it is possible to print out either 0, 1, 2 or 3 in the output?

[kenneth6]

The convenient, simple, and most often used method for figuring out the size of an array is...not to figure it out. Make it a separate variable, or if it's a function, require an additional argument indicating the size. Especially for the function, if someone gives you the wrong size, well, that's their own fault.

Yes, right ! Ganon11. I do agree with you. Avoid figuring it out!!

I tried to solve by adding a tracking variable!!

this is the most simple solution

[weaknessforcats]

OK. Here's how it works:

1) sizeof returns the s1ze of the object on the stack.

That makes:

[code=c]
char arr[5];
cout << sizeof(arr) << endl;
[/code]

display a value of 5. That is, arr occupies 5 bytes.

This:
[code=c]
char* arr = new char[5];
cout << sizeof(arr) << endl;
[/code]

displays 4. All that's on the stack is the pointer to the array.

As a rule, do not use sizeof to find the number of elements in an array.
1) It does not work for heap arrays
2) It does not work inside functions where the array is passed in.

Instead, use a const int var for the number of elements and pass that variable around alogn with the array address.