The member occupancies in the structure are generally not based on type size, and there is a so-called "alignment" problem. num occupies 4 bytes, name 20 bytes, sex 1 byte, ***25 bytes; the next age is not placed at the 26th byte, because it must be stored starting from an even number and it must be the longest type of integer times, you have to start storing age from the 29th byte. Calculated in this way, after storing addr, 66 bytes are needed; but 66 is not a multiple of the longest type 4, so in the end there are still 2 bytes left empty and the "alignment" is 68 bytes. That is to say, in this structure, 3 bytes are empty after sex and 2 bytes are empty at the end, so the total size is 68 bytes.
If you are not preparing for an exam, I suggest you just know that this is the case and don’t go into details...