addressable | alternative implementation to the URI implementation

Valgrind is giving me bizarre output that goes away if more memory is allocated. In my program, the number I have to add to make it go away is 2064. This number appears nowhere in my program, and I have been up and down the thing for a few hours now trying to find where I could have gone wrong. No luck. Everything seems airtight, and I see no reason why an extra 2064 bytes of memory should be needed.

Thankfully, I managed to reproduce the bug in a minimal program.

Im running ruby version 2.6.1 with docker. Rake gem is version 13.0.1.
Whenever I tried docker-compose up, it always fails and throws this error everytime:
This error did not exist before.

I am parsing a police records csv file. There are multiple occurrences of the same date in the file. I have the dates map that creates a new date key with the type of DateData assigned to it. The DateDate for now only has a Count key. That key is to be incremented every time the same date appears again in the loop. Right now I am only able to increment once. the count goes from 1 to 2 even if there are dozens of the same dates.

This is the output I get with a shorter version of the csv file for clarity purposes.

The following code is a sample of the project I'm currently working on, coded in C.

I first malloc a struct, and then as an example malloc the string inside the first one. When I try to copy text from another string into it, and print it using the printf function, it overflows when I compile using -fsanitize=address as compilation flag.

I don't understand why though, as I think I'm allocating enough memory to the string given I'm just taking the length using strlen of the other one, with one additional character for the \0.

I just started learning C++ and was surprised to see that when calling &varname, I got a 8 hex-digit-long number. Assuming this math is correct:

2^(8 hexits x 4 bits hexit^-1) x 1 bytes address^-1 = 4.29497e+09 bytes of RAM addressable

I should only be able to address 4GB of ram, and yet my computer has 16GB, all of which work. Does this mean my computer actually has 32 bits of storage per memory address?

I understand theory that is behind tagged pointers and how it is used to save additional data in pointer.
But i dont understand this part (from wikipedia article about tagged pointers).

Most architectures are byte-addressable (the smallest addressable unit is a byte), but certain types of data will often be aligned to the size of the data, often a word or multiple thereof. This discrepancy leaves a few of the least significant bits of the pointer unused

Why is this happening ?
Does pointer have only 30 bites (on 32 bit architectures) and that 2 bites are result of aligning?
Why there are 2 bites left unused in first place ?
And does this decrease size of addresable space (from 2^32 bytes to 2^30 bytes)?

I am trying to install a specific jq version in my virtual pyenv environment with the command below:

When I sudo gem install --verbose colorls, it succeeds.

I am reading up a bit on Master Boot Record layout and I was particularly interested in how the partition layout causes a size limitation on the size of the storage that can be used on a device with MBR.

Each partition within an MBR is defined using a 16 byte entry. The usage of those 16 bytes is as follows:

• 1st byte, if it has a value of 80, indicates active partition
• 2nd byte, the head number where the partition begins. This means MBR can address 256 different heads
• 3rd byte, the first 6 bits are used to capture the sector number of the 1st sector of the partition. This means MBR can address 64 different sectors
• 4th byte + last 2 bits of 3rd byte (total of 10 bits) store the track number where the partition begins. This means a total of 1024 tracks can be addressed using MBR partition entry.
• 5th byte (OS indicator)
• 6th byte the head number where the partition ends
• 7th byte, the first 6 bits are used to capture the sector number of the last sector of the partition
• 8th byte + last 2 bits of 7th byte store the track number where the partition ends
• Bytes 9, 10, 11, and 12 capture how many sectors where there before the beginning of the partition
• Bytes 13, 14, 15, and 16 capture how many sectors are there in the partition

Suppose we have only 1 partition in MBR and I make that the active partition. The zeroth sector is occupied by the MBR itself while the first partition starts from sector 1. Then the total number of sectors in this partition are: