YEAR 2038 PROBLEM (Computers)

The Year 2000 problem is understood by most people these days because of the large amount of media attention it received. Most programs written in the C programming language are relatively immune to the Y2K problem, but suffer instead from the Year 2038 problem. This problem arises because most C programs use a library of routines called the standard time library . This library establishes a standard 4-byte format for the storage of time values, and also provides a number of functions for converting, displaying and calculating time values.

The standard 4-byte format assumes that the beginning of time is January 1, 1970, at 12:00:00 a.m. This value is 0. Any time/date value is expressed as the number of seconds following that zero value. So the value 919642718 is 919,642,718 seconds past 12:00:00 a.m. on January 1, 1970, which is Sunday, February 21, 1999, at 16:18:38 Pacific time (U.S.). This is a convenient format because if you subtract any two values, what you get is a number of seconds that is the time difference between them. Then you can use other functions in the library to determine how many minutes/hours/days/months/years have passed between the two times.

First lets have brief introduction of bits and byte. Computers use binary numbers, and therefore use binary digits in place of decimal digits. The word bit is a shortening of the words "Binary digIT." Whereas decimal digits have 10 possible values ranging from 0 to 9, bits have only two possible values: 0 and 1. Therefore, a binary number is composed of only 0s and 1s, like this: 1011. How do you figure out what the value of the binary number 1011 is? We use a base of 2 instead of a base of 10. So:

(1 * 2^3) + (0 * 2^2) + (1 * 2^1) + (1 * 2^0) = 8 + 0 + 2 + 1 = 11

Bits are rarely seen alone in computers. They are almost always bundled together into 8-bit collections, and these collections are called bytes. Why are there 8 bits in a byte? A similar question is, "Why are there 12 eggs in a dozen?" The 8-bit byte is something that people settled on through trial and error over the past 50 years.

With 8 bits in a byte, you can represent 256 values ranging from 0 to 255, as shown here:

0 = 00000000

1 = 00000001

2 = 00000010

...

254 = 11111110

255 = 11111111

Same way a signed 4-byte integer has a maximum value of 2,147,483,647, and this is where the Year 2038 problem comes from. The maximum value of time before it rolls over to a negative (and invalid) value is 2,147,483,647, which translates into January 19, 2038. On this date, any C programs that use the standard time library will start to have problems with date calculations.

This problem is somewhat easier to fix than the Y2K problem on mainframes, Well-written programs can simply be recompiled with a new version of the library that uses, for example, 8-byte values for the storage format. This is possible because the library encapsulates the whole time activity with its own time types and functions (unlike most mainframe programs, which did not standardize their date formats or calculations). So the Year 2038 problem should not be nearly as hard to fix as the Y2K problem was.

Prepared by: Vishal Shah & Rasul Saiyad

U. V. Patel College Of Engineering