Big empty space in memory?

Question 1

The first two lines are the 8 interrupt vectors, expressed as 32-bit instructions with the highest byte last. That is, read them in groups of 4 bytes, with the highest byte last, and then convert to an instruction via the usual method. The first few vectors, including the reset at memory location 0, turn out to be LDR instructions, which load an immediate address into the PC register. This causes the processor to jump to that address. (The reset vector is also the first instruction to run when the device is switched on.)

You can see the structure of an LDR instruction here, or at many other places via an internet search. If we write the reset vector 18 f0 95 e5 as e5 95 f0 18, then we see that the PC register is loaded with the address located at an offset of 0x20.

So the next two lines are memory locations referred to by instructions in the first two lines. The reset vector sends the PC to 0x00000080, which is where the C runtime of your program starts. (The other vectors send the PC to 0x00000170 near the end of your program. What this instruction is is left to the reader.)

Typically, the C runtime is code added to the front of your program that loads the global variables into RAM from flash, and sets the uninitialized RAM to 0. Your program starts after that.

Your original question was: why have such a big gap of unused flash? The answer is that flash memory is not really at a premium, so we can waste a little, and that having extra space there allows for forward-compatibility. If we need to increase the vector table size, then we don't need to move the code around. In fact, this interrupt model has been changed in the new ARM Cortex processors anyway.

Question 2

Physical (not virtual) memory addresses map to physical circuits. The lowest addresses often map to registers, not RAM arrays. In the interest of consistency, a given address usually maps to the same functionality on different processors of the same family, and missing functionality appears as a small hole in the address mapping.

Furthermore, RAM is assigned to a contiguous address range, after all the I/O registers and housekeeping functions. This produces a big hole between all the registers and the RAM.

Alternately, as @Martin suggests, it may represent uninitialized and read-only Flash memory as -- bytes. Unlike truly unassigned addresses, access to this is unlikely to produce an exception, and you might even be able to make them "reappear" using appropriate Flash controller commands.

On a modern desktop-class machine, virtual memory hides all this from you, and even parts of the physical address map may be configurable. Many embedded-class processors allow configuration to the extent of specifying the location of the interrupt vector table.

Question 3

UncleO is right but here is some additional information.

The project's linker command file (*.icf for IAR EW) determines where sections are located in memory. (Look under Project->Options->Linker->Config to identify your linker configuration file.) If you view the linker command file with a text editor you may be able to identify where it locates a section named .intvec (or similar) at address 0x00000000. And then it may locate another section (maybe .text) at address 0x00000080.

You can also see these memory sections identified in the .map file, along with their locations. (Ensure "Generate linker map file" is checked under Project->Options->Linker->List.) The map file is an output from the build, however, and it's the linker command file that determines the locations.

So that space in memory is there because the linker command file instructed it to be that way. I'm not sure whether that space is necessary but it's certainly not a problem. You might be able to experiment with the linker command file and move that second section around. But the exception table (a.k.a. interrupt vector table) must be located at 0x00000000. And you'll want to ensure that the reset vector points to the new location of the startup code if you move it.