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If i got it right, Michelson is a stack based language.
I've read the following example about stack based languages:
3 4 5 * + = 23
how does Michelson know, that it has to mutiplicate 4 * 5 before adding 3?
Just to explain how i think:
The multiplicator comes first so i would think: 3*4+5 = 17
Why is that wrong?
We can simulate the program in the following way.
We write the stack between [...] and then the operations.
We use three different operations push <int>, * and +.
push puts an integer at the top of the stack, * pops two elements from the stack and pushes their product, and + pops two elements from the stack and pushes their sum.
Your program starts with an empty stack [] and some operations:
[], push 3, push 4, push 5, *, +
-> (evaluation of push 3)
[3], push 4, push 5, *, +
-> (evaluation of push 4)
[3, 4], push 5, *, +
-> (evaluation of push 5)
[3, 4, 5], *, +
-> (evaluation of *)
[3, 20], +
-> (evaluation of +)
[23]
So the result of your computation is 23.
'*' and '+', in this context, are not the infix operators you are used to. Imagine reading left to right and stacking up the numbers as you go. When you encounter + or * you replace the top two elements of the stack with the result of the operation.
