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({}()) What this does should be pretty clear. {} grabs a value from the stack, which implicitly zero to begin with, () adds one to it and (...) pushes the value. On the second run since there is already a 1 on the stack this just adds another 1 to it to make two. In fact if you copy the code n times it will always output n.
({}()) What this does should be pretty clear. {} grabs a value from the stack, which implicitly zero to begin with, () adds one to it and (...) pushes the value. On the second run since there is already a 1 on the stack this just adds another 1 to it to make two. In fact if you copy the code n times it will always output n.
({}()) What this does should be pretty clear. {} grabs a value from the stack, which implicitly zero to begin with, () adds one to it and (...) pushes the value. On the second run since there is already a 1 on the stack this just adds another 1 to it to make two. In fact if you copy the code n times it will always output n.
({}()) What this does should be pretty clear. {} grabs a value from the stack, which implicitly zero to begin with, () adds one to it and (...) pushes the value. On the second run since there is already a 1 on the stack this just adds another 1 to it to make two. In fact if you copy the code n times it will always output n.
({}()) What this does should be pretty clear. {} grabs a value from the stack, which implicitly zero to begin with, () adds one to it and (...) pushes the value. On the second run since there is already a 1 on the stack this just adds another 1 to it to make two. In fact if you copy the code n times it will always output n.
({}()) What this does should be pretty clear. {} grabs a value from the stack, which implicitly zero to begin with, () adds one to it and (...) pushes the value. On the second run since there is already a 1 on the stack this just adds another 1 to it to make two. In fact if you copy the code n times it will always output n.
({}()) What this does should be pretty clear. {} grabs a value from the stack, which implicitly zero to begin with, () adds one to it and (...) pushes the value. On the second run since there is already a 1 on the stack this just adds another 1 to it to make two. In fact if you copy the code n times it will always output n.
({}()) What this does should be pretty clear. {} grabs a value from the stack, which implicitly zero to begin with, () adds one to it and (...) pushes the value. On the second run since there is already a 1 on the stack this just adds another 1 to it to make two. In fact if you copy the code n times it will always output n.