we were given a python bytecode and asked to figure out the equivalent code in python. at first, it wasn't easy for me to understand, but I got over it.
So here is the bytecode, and what it means
3 0 LOAD_CONST 1 (98) 3 LOAD_FAST 0 (a) 6 LOAD_FAST 1 (b) 9 BINARY_POWER 10 BINARY_ADD 11 RETURN_VALUE
3(number at the first column) means the current line of execution in the source code.
second columnn contains the offset of address of the instuctions
third column is the opname for the instructions(they are just meaning/human readeable names given to instructions)
fourth column is the index of the arguments to the instructions
fifth column is interpretation of the parameters in parentheses.(they are not of the original bytecode, i think this is added by the dis() to help humans understand what the bytecode is doing
LOAD_CONST loads a constant value onto the stack, in this case, the constant 1 with a value of 98.
LOAD_FAST loads the value of the local variable a onto the stack.
LOAD_FAST loads the value of the local variable b onto the stack.
BINARY_POWER performs the exponentiation operation between the top two values on the stack. It raises the value of b to the power of a in this case.
BINARY_ADD adds the top two values on the stack. It adds the result of the exponentiation operation (from line 9) with the constant 98 (from line 0).
RETURN_VALUE returns the value on top of the stack, which is the result of the addition operation. In summary, this code snippet calculates the result of (a ** b) + 98 and returns the final value. The operator ** represents exponentiation, so it's raising the value of b to the power of a. The constant 98 is then added to the result.
more on what the fourth column means
for example in this line
3 0 LOAD_CONST 1 (98)
In the bytecode instruction LOAD_CONST, the argument specifies the index of a constant in the constant table of the code object. The constant table is a list of values used within the code block.
The instruction LOAD_CONST is used to load a constant onto the stack. The argument 1 indicates that the constant being loaded is at index 1 in the constant table. In this case, the constant value is 98.
The constant table is an internal data structure that stores values such as integers, strings, and other immutable objects that are used within the code block. The bytecode instructions then refer to these constants by their index in the constant table.
To summarize, the argument 1 in the LOAD_CONST instruction specifies the index of the constant 98 within the constant table of the code object.
another example, with the second line
3 LOAD_FAST 0 (a)
The bytecode instruction LOAD_FAST is used to load the value of a local variable onto the stack. The argument 0 in this instruction indicates the index or position of the local variable within the local variables table.
The local variables table is an internal data structure that stores the local variables of a function or code block. The index of a local variable represents its position within that table.
In this case, the instruction LOAD_FAST is loading the value of the local variable a onto the stack. The argument 0 indicates that a is the first local variable in the local variables table.
Once the value of a is loaded onto the stack, it can be used for further operations or computations in the code.
Now let me explain the stack and its order in this context
this piece was quite confusing for me, had to research more to come up with a pattern(works for now)
code is code everywhere, in every language, asm, bytecode they are executed line by line(at blazing speed rather)
first instr 98 is loaded on top of the stack -> TOS2
second instr a is loaded on top of 98 -> TOS1
third instr b is loaded on top a -> TOS(Top of stack)
fourth instr BINARY_POWER: takes two first values from the stack TOS and TOS1 and performs this operation a**b and save it back to the stack
fifth instr BINARY_ADD: takes two values and performs this operation 98 + (a**b)
note the pattern and order i'm taking(if you want to accurately get the source code)
first value popped from the stack goes to the right, the command in the middle and the second value popped goes to the left just like in a**b
b to right
a to left
and in 98 + (ab) ab to the left 98 to the right