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GoFiler Legato Script Reference
Legato v 1.5b Application v 5.24b
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Chapter Three — Data Types and Operators (continued)
The order of evaluation of operators is key in determining the result of an expression and it becomes more important as expressions grow in length and complexity. The following table shows the order of operator precedence in Legato:
| Precedence | Operator | Description | Associativity | |||||
| 1 (Highest) | ++ | Suffix increment | Left-to-right | |||||
| -- | Suffix decrement | Left-to-right | ||||||
| () | Function call, precedence group. | Left-to-right | ||||||
| [] | Array subscript | Left-to-right | ||||||
| 2 | ++ | Prefix increment | Right-to-left | |||||
| -- | Prefix decrement | Right-to-left | ||||||
| + | Unary plus (positive) | Right-to-left | ||||||
| - | Unary minus (negative) | Right-to-left | ||||||
| ! | Logical NOT | Right-to-left | ||||||
| ~ | Bitwise NOT (One’s Complement) | Right-to-left | ||||||
| 3 | * | Multiplication | Left-to-Right | |||||
| / | Division | Left-to-Right | ||||||
| % | Modulo | Left-to-Right | ||||||
| 4 | + | Addition | Left-to-Right | |||||
| - | Subtraction | Left-to-Right | ||||||
| 5 | >> | Bitwise left shift | Left-to-Right | |||||
| << | Bitwise right shift | Left-to-Right | ||||||
| 6 | < | Less than | Left-to-Right | |||||
| <= | Less than or equal to | Left-to-Right | ||||||
| > | Greater than | Left-to-Right | ||||||
| => | Greater than or equal to | Left-to-Right | ||||||
| 7 | == | Equal to | Left-to-Right | |||||
| != | Not equal to | Left-to-Right | ||||||
| 8 | & | Bitwise AND | Left-to-Right | |||||
| 9 | ^ | Bitwise XOR | Left-to-Right | |||||
| 10 | | | Bitwise OR | Left-to-Right | |||||
| 11 | && | Logical AND | Left-to-Right | |||||
| 12 | || | Logical OR | Left-to-Right | |||||
| 13 | ?: | Ternary conditional | Right-to-Left | |||||
| 14 | = | Direct assignment | Right-to-Left | |||||
| += | Assignment by sum | Right-to-Left | ||||||
| -= | Assignment by difference | Right-to-Left | ||||||
| *= | Assignment by product | Right-to-Left | ||||||
| /= | Assignment by quotient | Right-to-Left | ||||||
| %= | Assignment by remainder | Right-to-Left | ||||||
| <<= | Assignment by bitwise left shift | Right-to-Left | ||||||
| >>= | Assignment by bitwise right shift | Right-to-Left | ||||||
| &= | Assignment by bitwise AND | Right-to-Left | ||||||
| ^= | Assignment by bitwise XOR | Right-to-Left | ||||||
| |= | Assignment by bitwise OR | Right-to-Left | ||||||
| 15 (Lowest) | , | Comma | Left-to-Right |
3.12.2 A Note About Interpreted Versus Compiled
In the Legato interpreted environment, the engine creates a list by precedence of items to perform and then performs them in order. For example:
if ((a < b) || (b < c)) { something ... }
In this case, the operations are broken down as follows:
a < b to result 1
b < c to result 2
result 1 || result 2 to result 3
if (...) result 3 then { something ...}
The parentheses in the above case are superfluous but can help with reading clarity. In an optimized compiler environment, the second operation, b < c will never execute if the first condition is true, it is logically unnecessary. However, this leads to some interesting differences. For example:
if ((a < b[i++]) || (b[i++] < c)) { something ... }
In this case, the operations are broken down as follows:
a < b[i] to result 1
++ (i = i + 1)
b[i] < c to result 2
++ (i = i + 1)
result 1 || result 2 to result 3
if (...) result 3 then { something ...}
Two conditions happen in an optimized compiler. First, i placed in a temporary area so every reference to i and a suffix unary condition will be based on the original value prior to the execution of the expression. Second, and more of a logical oddity, if the second operation, b < c, is not executed, the value of i will be incremented only once.
In Legato, the operations are executed in the order as above. So, i is 0 on start. It will always end as 2. Further, result 2 is based on b[1], not b[0]. In general, the second example is simply bad form and should be avoided.
Another odd example exposing the same unary differences:
while (a < 5) { list[a++] = a; }
The list in this condition will be loaded with 1,2,3,4,5, not 0,1,2,3,4. Again, this practice should be avoided.
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