How are the maximum single jump range of MIPS branch instructions calculated?

Question

I am studying computer architecture (MIPS architecture) and read the following statements:

1.Branch instructions have a 16 bit signed word offset field that allows a branch to an address + or -128kBytes (+0x1FFFC TO -0X20000) from the current location.

2.A jump instruction specifies an address within the current 256MByte(0x0FFFFFFC) region specified by Program counter most significant 4 bits.

I understand the concept of jump range described above but how are the three numbers 0x0FFFFFFC, 0x1FFFC and 0X20000 calculated using "the range of 256Mbyte" and "the range of +-128 kbytes"?

Thanks!

1 word = 4 bytes. So obviously a 16 bit signed word offset is +/- 128k and I hope you have no problem converting that into hex. As for the 256MByte that is not an offset. So the biggest jump you can make is of course 256M-4. — Jester
Related: How to Calculate Jump Target Address and Branch Target Address? shows the encoding. — Peter Cordes

adnaan adnaan · Accepted Answer · 2021-02-13T17:51:21

The other answers didn't really answer your question of how these Hex values are calculated/found. so here's my answer.

Thinking about this is much easier in Binary than HEX. as the 2bit left shift is important to understanding the concept 2bits is multiply by 4. Which cannot be represented in HEX as nicely since easy Hex digit is 16 values. but ill try to explain it still:

0x20000

1 Branch instructions use a 16 bit Immediate field. (5 bit RS, RT) (6 bit Opcode) == 32bits (https://en.wikibooks.org/wiki/MIPS_Assembly/Instruction_Formats#I_Format)

those 16 bits are Signed. they can be positive & Negative.

That gives you an effective range of -(2^15) == -32768

to +(2^15 -1) == 32767

MIPS multiples any address inputs by 4. Forcing them to be word aligned.

so your Minimum value -(2^15) Multiply by 4: -{2^15 *4} (4=2^2), {2^(15+2)} (15+2 == 17): becomes -(2^17) == -131072

in Binary (signed 2's complement). 1000 0000 0000 0000 <<2 == 10 0000 0000 0000 00[00]

Converting that to Hex 10=2 (0000=0) gives 2 0 0 0 0 == 0x20000

this would be sign extended before adding it to the (PC+4):

so for say, instruction #32770, PC=0x00420008 (PC+4)=0x0042 000C

0x0042000C - 0x20000 = 0x0040000C, instruction #3 (remember, offset is based off PC+4)

#32770+1 +-32768 == 3

0x1FFFC

Same for the Maximum value: (2^15 -1) Multiply by 4: {(2^15 -1) *4} (4=2^2), {2^(15+2) -(1*4)} (15+2 == 17):

becomes (2^17 -4) == 131068 0111 1111 1111 1111 <<2 == 01 1111 1111 1111 11[00]

Converting that to Hex 01=1 (1111=F) (1100=C) gives 1 F F F C == 0x1FFFC

Note the address needs to be added to the current (Program Counter+4)

so for say, instruction #32770, PC=0x00420008 (PC+4)=0x0042000C

0x0042000C + 0x1FFFC= 0x440008, instruction #65538 (remember, offset is based off PC+4)

#32770+1 +32767 == 65538

0x0FFFFFFC

2 now Jumps use a 28 bit address.
Also Note, Jumps use an absolute address. not an offset.

maximum 28 bit value is (2^26 -1) == 67108863, 0x03FFFFFF ``

Shifted 2 (*4) becoming 28bits. {(2^26 -1) *4}, == {2^28 -4} == 268435452, 0x0FFFFFFC

But then the missing four bits ? .. they come from the PC - which in the Memory stage, it has already been incremented to (PC+4)

for instruction #32770, PC=0x00420008 (PC+4)=0x0042000C 0x0042000C in binary is [0000] 0000 0100 0010 0000 0000 0000 1100

+0x0FFFFFFC in binary [####] 1111 1111 1111 1111 1111 1111 1100 it is only 28 (27:0) bits and missing the 31:28 bits.

Taking the bits from PC+4. we get:

 0000 ---- ---- ---- ---- ---- ---- ---- (PC+4)
 ---- 1111 1111 1111 1111 1111 1111 1100 (Target-Address)
-----------------------------------------
 0000 1111 1111 1111 1111 1111 1111 1100 (Jump-Address)

(which in this case is the same value as sign extending it)

A better explanation of how Addresses are calculated. How to Calculate Jump Target Address and Branch Target Address?

How are the maximum single jump range of MIPS branch instructions calculated?

2 Answers

0x20000

0x1FFFC

0x0FFFFFFC