Beginners: Did you know assignment operators can do this

[TBMan]

I just saw this thread and I tried...

Code: Pascal  [Select][+][-]

1. var
2. x:integer;
3. begin
4. x := 6;
5.  
6. x+=1;
7.  
8. writeln(x);
9.  
10. end.
11.  

And 7 was the output. I like that.  I was always jealous of the C syntax  X++;

Doing the following is self documenting though:

Code: Pascal  [Select][+][-]

1.  
2. Inc(x);  // if you know Pascal you know.  It's a procedural call though just to add 1 to a number, and
3.  
4. Inc(x,#)   // there is this too.          
5.  
6. // or
7.  
8. x := x+1;  // best thing for readability, but always felt clumsy
9.  
10.  

[Khrys]

[...] plays games with expected Pascal precedence [...]

Expected Pascal precedence... more like unexpected Pascal precedence 

Code: Pascal  [Select][+][-]

1. if not Number in [2, 3, 5, 7, 11] then {...}

if  statements don't require parentheses per se, but don't worry - you're gonna need lots of them regardless:

Code: Pascal  [Select][+][-]

1. if X = 0 and I < N - 1       then { Error: Incompatible types: got "Boolean" expected "Int64" - Very helpful! }
2. if (X = 0) and (I < (N - 1)) then { Whoops - got too paranoid there, (N - 1) doesn't need parentheses }
3. if (X = 0) and (I < N - 1)   then {...}

[Thaddy]

I know the precedence rules.
 It should be evaluated as x := x * y + 1 because that is what it shortcuts. Not x:= x * (y+1). Anyway, I am used to it, but that is not Pascal.

[Leledumbo]

// if you know Pascal you know.  It's a procedural call though just to add 1 to a number, and

Thankfully there's no mandate to make that a procedure call.

Code: Pascal  [Select][+][-]

1. Inc(x);
2. Inc(x,#);
3.  

Is compiled by fpc, even without any switches, to:

Code: ASM  [Select][+][-]

1. addl    $1,-8(%rbp)
2. addl    $5,-8(%rbp)
3.  

and with -O2:

Code: ASM  [Select][+][-]

1. addl    $6,%edi
2.  

The advantage of it being a compilerproc

[Thaddy]

procedural call <> procedure call. The former also describes compilerproc