Does the stack in this 'ld' linker script overwrite the stored executable?

I have a question about the behavior of the linker script found in this question:

https://stackoverflow.com/a/55193198/2421349

To save you a click, the relavant portion is:

OUTPUT_ARCH(riscv)
MEMORY
{
   /* qemu-system-risc64 virt machine */
   RAM (rwx)  : ORIGIN = 0x80000000, LENGTH = 128M 
}
ENTRY(_start)

And in a later section:

 PROVIDE (__executable_start = SEGMENT_START("text-segment", ORIGIN(RAM)));
 . = SEGMENT_START("text-segment", ORIGIN(RAM)) + SIZEOF_HEADERS;
 PROVIDE(__stack_top = ORIGIN(RAM) + LENGTH(RAM));

We set __executable_start to begin at ORIGIN(RAM). Then we use the . command to move the linker output location SIZEOF_HEADERS bytes forward. And finally we set __stack_top = ORIGIN(RAM) + LENGTH(RAM).

Assuming the stack grows down towards ORIGIN(RAM), won't it eventually overwrite __executable_start and whatever SIZEOF_HEADERS is if the stack grows large enough?

Yes, if the stack grows large enough, it will eventually start overwriting parts of the memory it should not. But this is not specific to this linker script: Ultimately, the memory is a finite resource, and any stack growing too much because too large automatic variables beeing allocated, and/or out of control recursive calls will end up causing problems.