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Unix time memory problem

Already answered- thanks
Jan 14 '20 #1

✓ answered by SioSio

In traditional UNIX implementations, time_t was an int, and the int was signed 32 bits. The count starts with 0:00:00 on January 1, 1970, universal time, as 0 seconds.
Therefore, the maximum value is (2^31−1) = 2,147,483,647, and it can handle 2,147,483,647 seconds (≒ 68 years, until 3:14:07 on January 19, 2038).
If the time_t type is changed to a signed 64-bit integer type, the maximum value is (2^63-1) = 9,223,372,036,854,775,807, and can be used up to 300 billion years in the Christian era.

Based on this, it is 36 bits to be able to count until the end of 3038.

2 2459
SioSio
272 256MB
In traditional UNIX implementations, time_t was an int, and the int was signed 32 bits. The count starts with 0:00:00 on January 1, 1970, universal time, as 0 seconds.
Therefore, the maximum value is (2^31−1) = 2,147,483,647, and it can handle 2,147,483,647 seconds (≒ 68 years, until 3:14:07 on January 19, 2038).
If the time_t type is changed to a signed 64-bit integer type, the maximum value is (2^63-1) = 9,223,372,036,854,775,807, and can be used up to 300 billion years in the Christian era.

Based on this, it is 36 bits to be able to count until the end of 3038.
Jan 15 '20 #2
dev7060
633 Expert 512MB
The whole thread doesn't make sense without the original post. It isn't appropriate to remove the question. The discussion may benefit others as well. Same thing noticed here.
Jan 15 '20 #3

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