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I was just disassembling and debugging an ARM binary for fun and I noticed something unusual. Consider the following set of instructions:-

   0x00008058 <+4>: mov r1, pc
   0x0000805c <+8>: add r1, r1, #24
   0x00008060 <+12>:    mov r0, #1

I tried setting a breakpoint at 0x0000805c and checked the value of the register r1. I was expecting to see 0x0000805c -- however, interestingly the value is 0x8060.

Why is this? Is this because of some sort of instruction pipelineing?

jvoisin
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1 Answers1

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Yes, it's because of pipelining.

From http://winarm.scienceprog.com/arm-mcu-types/how-does-arm7-pipelining-works.html --

ARM pipelining

PC (Program Counter) is calculated 8 bytes ahead of current instruction.

Jason Geffner
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