In csh, why is 4 - 3 + 1 == 0?

Question

#!/bin/csh @ cows = 4 - 3 + 1 echo $cows 

This simple csh script on startup creates a "0" for output when I expect a "2".

 ~root: csh simple.1 0 

I made a bunch of looks, and the only thing I could think of was that the “-” was read as a unary negation, not a subtraction, so changing the operator’s priority and ending with 4-4, not 2 + 1. Is that right? If so, why? If not ... help!

Edit: that means they are associative right! These operators are NOT associative in C, right? Is C-Shell different from C?

Answer 1

As long as you expect the operators to remain associative, they are right associative in csh , so it evaluates to 4-(3+1)

  - / \ / \ 4 + / \ 3 1 

Answer 2

The + and - operators are right-associative in csh. This means that "4 - 3 + 1" is rated as "4 - (3 + 1)".

Answer 3

Grouping of operators. It reads the operation as 4 - (3 + 1), unlike (4 - 3) + 1.