# how pseudo direct addressing works?

In pseudo direct addressing mode (For the MIPS architecture) the 26 bit of the jump instruction are joined to the upper 4 bits of the PC .

• how could this help in jumping to relative positions suppose I want to jump backward instead of forward ( i.e. if i want to jump to the beginning of a loop after executing it's body ),

• What I get is this will limit the jump to be 2^28 from the following instruction but Is this applicable backward too ? I am really confused by this so excuse me for such a silly question...

• I think how addressing works will depend on the specific architecture. Which CPU architecture are you studying? Please edit your question to provide a little bit of context.
– D.W.
Mar 26, 2020 at 19:31
• @D.W. sorry I just thought it's the same , I am reading the 'computer organization and design ...' in which it uses the mips architecture. Mar 27, 2020 at 17:11

Pseudo direct addressing is not intended for relative jumps. It is an absolute jump. If you want a relative jump, don't use pseudo direct addressing.

It's not correct that the limit is $$2^{28}$$ past the current address. Suppose the current address is 0x183020C0 (to give an example). Then pseudo direct addressing will allow jumping to any address in the range 0x10000000 - 0x1FFFFFFC. Notice that this is up to 0x083020C0 backwards and up to 0xF7CFDF3C forwards from where you currently are; but crucially, those values depend on the current PC. This is not a PC-relative jump. It is a jump to a fixed, absolute constant address.

Typically the code segment will fit within 256MB, so this is typically sufficient to jump to any (fixed) instruction within the code segment.

• what if the current PC value is already 0x10000000 this way you can't jump any backward this is relative to what PC is pointing at . right ? Mar 27, 2020 at 17:54
• @MahmoudSalah, yes, then you can only jump to instructions that are later, not earlier.
– D.W.
Mar 27, 2020 at 18:07

In MIPS-speak, these areas of memory of 28 (26+2; the +2 is because instructions are four bytes in size) bit address space (256MB) are referred to as "superblocks". Pseudo-direct addressing allows you to jump anywhere within the same superblock. It doesn't matter if it's forwards or backwards.

This was never a problem in practice because it's very rare for a single binary (whether it's an executable or a shared library) to have more than 256MB of code. IIRC, the MIPS linker could handle it if it ever occurred by generating a trampoline. Suppose that the compiler generated a jal, for example, where the linker discovered that the target was in a different superblock. Then the linker would generate a couple of instructions to do the full jump, and point the jal to that.

• still didn't get how it can jump backward if it just concatenating the 26 bit to the first 4 bits of the PC i.e. could you please give an example of a value in PC which will be modified to jump backward ? Mar 27, 2020 at 17:17
• Imagine the instruction that you're jumping from is the last one in the superblock. Mar 27, 2020 at 21:25