C++ Array Size

Get the Size of an Array

To get the size of an array, you can use the sizeof() operator:

int myNumbers[5] = {10, 20, 30, 40, 50};
cout << sizeof(myNumbers);

20

Why did the result show 20 instead of
5, when the array contains 5 elements?

It is because the sizeof() operator returns the size of a type in bytes.

You learned from the Data Types article that an int type is usually 4 bytes, so from the example above, 4 x 5 (4 bytes x 5 elements) = 20 bytes.

To find out how many elements an array has, you have to divide the size of the array by the size of the first element in the array:

int myNumbers[5] = {10, 20, 30, 40, 50};
int getArrayLength = sizeof(myNumbers) / sizeof(myNumbers[0]);
cout << getArrayLength;

5

Loop Through an Array with sizeof()

In the Arrays and Loops article, we wrote the size of the array in the loop condition (i <
5). This is not ideal, since it will only work for arrays of a specified size.

However, by using the sizeof() approach from the example above, we can now make loops that work for arrays of any size, which is more sustainable.

Instead of writing:

int myNumbers[5] = {10, 20, 30, 40, 50};
for (int i = 0; i < 5; i++) {
  cout << myNumbers[i] << "\n";
}

It is better to write:

int myNumbers[5] = {10, 20, 30, 40, 50};
for (int i = 0; i < sizeof(myNumbers) / sizeof(myNumbers[0]); i++) {
  cout << myNumbers[i] << "\n";
}

Note that, in C++ version 11 (2011), you can also use the “for-each” loop, which is even cleaner and simpler:

int myNumbers[5] = {10, 20, 30, 40, 50};
for (int i : myNumbers) {
  cout << i << "\n";
}