I try to add new instruction,

// old

const/v4 v?, 0x0

// new

const/v4 v?, 0x0
const-string v?, "Hello" // new instruction

can someone help me, adding instruction without duplicate register number?

Look to these sample I've created DEXPatcher:
https://gitlab.com/CodeCracker/DEXPatcher
https://github.com/CodeCrackerSND/DEXPatcher
https://bitbucket.org/CodeCrackerSND/dexpatcher

More exactly look here:
https://gitlab.com/CodeCracker/DEXPatcher/blob/master/dexpatcher/DEXPatcher.java

 

Edited September 20, 20186 yr by CodeExplorer

@CodeExplorer

I try with this code:

private static MethodImplementation addNewInstructions(@Nonnull MethodImplementation implementation) {

		MutableMethodImplementation mutableImplementation = new MutableMethodImplementation(implementation);

		List<BuilderInstruction> instructions = mutableImplementation.getInstructions();
		
		for (int i = 0; i < instructions.size(); i++) {
			
			Instruction instruction = instructions.get(i);
			
			if (instruction instanceof ReferenceInstruction) {

				if (((ReferenceInstruction) instruction).getReferenceType() == ReferenceType.METHOD) {

					Opcode opcode = instruction.getOpcode();

					if (!opcode.name.endsWith("range")) {
						mutableImplementation.addInstruction(i++, new BuilderInstruction21c(Opcode.CONST_STRING,
								0, new ImmutableStringReference("Hello World"))); // look
					}
				}
			}
		}

		return mutableImplementation;
	}

after :

// before

.line 54
iget-object v0, p0, Lapp/test/MainActivity;->ed:Landroid/widget/TextView;


// after

.line 54
iget-object v0, p0, Lapp/test/MainActivity;->ed:Landroid/widget/TextView; // v0 already register

const-string v0, "Hello World" // v0 duplicate register

If i storing a new value in it, I'll be clobbering any value that was already there, and later code that expects the previous value will likely fail.

Edited September 20, 20186 yr by Modify