CODEWITHSUNDEEP
androiddalviksmali
Kikapi
Kikapi

Reputation: 369

understanding invoke-kind/range argument word count

I'm having trouble understanding the structure of invoke-kind/range opcode,

Syntax

invoke-kind/range {vCCCC .. vNNNN}, meth@BBBB

Arguments

A: argument word count (8 bits)

B: method reference index (16 bits)

C:first argument register (16 bits)

N = A + C - 1

As you can see B and C are mentioned in the bytecode syntax but A is not mentioned, Where is the A argument located and what it means exactly?

Thanks.

Upvotes: 4

Views: 582

Answers (1)

JesusFreke
JesusFreke

Reputation: 20282

A contains the number of registers that is being passed to the method.

So if you have invoke-static/range {v0 .. v7}, method, then A will be 8, and C is 0. You can see from the formula at the bottom that N, the last register being passed, is calculated as N = A + C - 1, so N = 0 + 8 - 1 = 7

Upvotes: 2

Related Questions