Why is the .bss segment required?

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The reason is to reduce program size. Imagine that your C program runs on an embedded system, where the code and all constants are saved in true ROM (flash memory). In such systems, an initial “copy-down” must be executed to set all static storage duration objects, before main() is called. It will typically go like this pseudo:

for(i=0; i<all_explicitly_initialized_objects; i++)
{
  .data[i] = init_value[i];
}

memset(.bss, 
       0, 
       all_implicitly_initialized_objects);

Where .data and .bss are stored in RAM, but init_value is stored in ROM. If it had been one segment, then the ROM had to be filled up with a lot of zeroes, increasing ROM size significantly.

RAM-based executables work similarly, though of course they have no true ROM.

Also, memset is likely some very efficient inline assembler, meaning that the startup copy-down can be executed faster.

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