This following is a simple, yet informative, exercise that provides insight into just how big a 64-bit integer truly is.
The maximum positive value for bigint is ((2^63)-1), and it requires 8 bytes of storage for each value. Using some very simple math we come up with the following
(Maximum positive bigint value) * (8 Bytes per value) = Total storage required to store all positive bigint values
((2^63)-1) * 8 = 73,786,976,294,838,206,456 Bytes
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73.7 EB (Exabyte’s)
65,536 PB (Petabytes)
67,108,864 TB (Terabytes)
That’s right, sixty-seven million Terabytes! This is without any additional data in the records, only a bigint ID column with all positive values. If you use negative values, then just double everything shown above.
Note: This same exercise, using the int (4 byte) data type, takes only 8 GB.