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ThePhantomUpvoter wrote: The -- is short hand for y = y - 1
That's not entirely correct.
The point behind the '--' being in front of the expression or behind it just tells the compile WHEN to increment or decrement the expression. The operation is either going to happen before the expression is evaluated, in his case y, or after.
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WidmarkRob wrote: Dummy it down for me quite a bit please.
OK, let's break it down.
WidmarkRob wrote: int x = 10;
x is now 10.
WidmarkRob wrote: int y = 100;
y is now 100. Pretty obvious so far.
But this
WidmarkRob wrote: int z = y-- + x;
is a weird thing, and in order to understand it, it is necessary to understand the difference between the result of an expression and the effect of an expression. The C# rules say this about post-decrement: "the operand is decremented. The result of the expression is the value the operand had before the decrement". (see operator --[^])
So what's happening here is that y is decremented (that is the effect), so after y-- has been evaluated y will be 99, but the value is the old y, 100 (that is the result). Then it goes on to add x to that result, no surprises there.
WidmarkRob wrote: z = --z + x;
This is different, a pre-decrement as it is called. The value of a pre-decrement is the new value, ie after the decrement.
ps: most of this applies to most C-derived languages, except that C# is a bit special in that it specifies that the side-effects happen from left to right. That only matters if you decrement or increment something, and then use that same thing again in the same expression (ok not really, it's more complex, but it's True Enough)
modified 28-Apr-13 7:54am.
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Okay, I think I got it now.
When the decrementing comes after the variable.
We evaluate the expression before the decrementing is applied.
Then whenever we use the old declared variable after that first evaluated expression, the decrementing is applied.
Is that right?
<a href="http://www.widmarkrob.com">My Coding Journey</a>
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Close enough. The decrementing itself really happens while that expression is being evaluated, but that rarely matters. It does matter if you do this:
int x = 0;
int z = x++ + ++x; After that, clearly x is 2, because it was incremented twice.
And z is also 2, proving that the increment really happens while the expression is evaluated. First x++ is evaluated, the result is 0 and the effect is that x is now 1. Then ++x is evaluated, x was 1 and for pre-increment the result is the value after the increment, so the result is two (the effect is again that x is incremented, of course). So that works out to 0 + 2.
The left-to-right rules goes even further, if you have an assignment where there's an increment on the left hand side (yes, you can do that), it happens before the right hand side is evaluated, which can be demonstrated with something like this:
int[] array = new int[2];
int x = 0;
array[x++] = x++; After that, x is 2, obviously, and array looks like this: { 1, 0 }
What happened, was the first x++ was evaluated first, making the index into the array 0, then the right hand side is evaluated, which is 1 due to the first increment, and then x is incremented again, and finally the value of the right hand side (which was 1) is stored in the array at index 0.
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Because I'm Just Beginning, I'm going to stop at "Close Enough".
I'm sure the book will eventually got into more details, but… For now, as long as they understand that basics.
I think I can move on.
Thank you for helping me understand this more, actually way mor. 
<a href="http://www.widmarkrob.com">My Coding Journey</a>
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after all that dumbing it down, I'm thrown for a loop (pun intended).
After reading a little bit more in the book they decide to give me a little exercise.
int x = 10;
int y = 100;
int z = y;
y = y++ + x;
z = ++z + x;
I thought I would do this in my head before writing it to the console window.
The first expression, the answer I got my head was:
110
(which turned out to be right, woo hoo)
The second expression I got wrong, in my head I came up with:
121
the console window printed out:
111
Console.WriteLine(y);
Console.WriteLine(z);
just before this exercise, the book showed me a table trying to explain:
primary, urnary, binary
<a href="http://www.widmarkrob.com">My Coding Journey</a>
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WidmarkRob wrote: int z = y;
Ok, what happens here is not "make x an other word for y", which is what I think you might have thought. It really means "let x have the value that y now has".
y later changed, z did not.
Ok z did change, but the assignment to y did not affect it.
WidmarkRob wrote: y = y++ + x;
does not affect z, only y.
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Try an experiment.
When you see a statement involving prefix or postfix increments, mentally (or even physically) re-write it to be several statements, moving the increment outside all other statements.
y = y++ + x;
Becomes:
int y2 = y;
y = y + 1;
y = y2 + x; In other words, the postfix increment of y is irrelevant, because the value is immediately discarded, and y is set to the value of the sum of the original value of y and x
z = ++z + x; Becomes:
z = z + 1;
z = z + x; That is all the compiler is doing - a prefix or suffix increment just gets done when it is met, that's all - it's syntactic sugar for the broken down statements above.
Having said that, try not to use them in "complex" statements: What happens may not be what you expect - different compilers interpret "when to do this" slightly differently, and that gave have a dramatic effect. Normally, pre-and post- increments are kept to simple things like array accesses and for loops.
The universe is composed of electrons, neutrons, protons and......morons. (ThePhantomUpvoter)
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OriginalGriff wrote: different compilers interpret "when to do this" slightly differently
Not in C#. Or if they do, it's a compiler bug. There is only one right order of side-effects in C#, which is left-to-right.
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I originally did add something to that effect, but deleted it because I didn't want to confuse the OP too much. Anyone who starts writing code like
int y = 10;
y = y++ + ++y*y--; deserves the headache they are going to get1.
1 When I find out and hit them.
154, by the way.
The universe is composed of electrons, neutrons, protons and......morons. (ThePhantomUpvoter)
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Ok, but IMHO you could now be confusing the OP with the possibility that there's some sort of quantum-magic involved that makes the result depend on the phase of the moon or something..
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Do you want to sit there and work out why y++ + ++y * y-- equals 154? Quantum magic works for me! 
The universe is composed of electrons, neutrons, protons and......morons. (ThePhantomUpvoter)
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Easy, 10 + 12 * 12. The complex cases don't really matter of course, I'm sure we're in agreement there - I just brought them into this to prove to OP that it really works that way.
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I know - but it's not that obvious when you look at it, particularly if you come from a C / C++ background where a different result is a strong possibility. DevC++ will give you 132 for example.
The universe is composed of electrons, neutrons, protons and......morons. (ThePhantomUpvoter)
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Also fun: different version of GCC give different results (usually 132 or 142).
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Now you see why I hit people who do it!
The universe is composed of electrons, neutrons, protons and......morons. (ThePhantomUpvoter)
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.. And if they turn the other cheek, hit them again This has got to be close to using GoTo statments or possibly worse....
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I think of it as "Codefuscating" 
The universe is composed of electrons, neutrons, protons and......morons. (ThePhantomUpvoter)
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To you or the compiler or perhaps both?
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Oh just me, the compiler knows what it's doing. I just guess. 
The universe is composed of electrons, neutrons, protons and......morons. (ThePhantomUpvoter)
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Quote: I just guess. debug.
FTFY
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Worse. With goto at least only one thing can happen, even if it's not always immediately clear what that thing is.
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I have seen some pretty nasty goto's in my life, but I have to agree that this seems to take the cake. Why would anybody want to use such a feature anyway, it seems daft.
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I see you guys have been busy bees about this… I had to take a break, steam was coming from my ears because of this.
Anyway, I looked over the example in the exercise they gave me.
Give me about 30 min., I'll explain to you guys but I ended up getting the second expression in my exercise wrong.
By the way, thank you for trying to help clear this up for me.
Rob
<a href="http://www.widmarkrob.com">My Coding Journey</a>
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Hi,
Sometimes a simple description can be very hard to grasp - especially if the simple description tries to abstract the details away.
In such cases I find it rewarding simply to dig a bit deeper...
Here goes the details:
.entrypoint
.maxstack 3
.locals init ([0] int32 x,
[1] int32 y,
[2] int32 z)
IL_0000: nop
IL_0001: ldc.i4.s 10
IL_0003: stloc.0
IL_0004: ldc.i4.s 100
IL_0006: stloc.1
IL_0007: ldloc.1
IL_0008: dup
IL_0009: ldc.i4.1
IL_000a: sub
IL_000b: stloc.1
IL_000c: ldloc.0
IL_000d: add
IL_000e: stloc.2
If instruction, stack and arithmic unit are alien terms to you, then above might be a bit tough. It is a window into a lower layer of code. The C# compiler outputs this in binary form as your assembly/executable. IL means Intermediate Language and it can be executed by a Virtual Machine aka a program. The idea is to simulate the arithmic unit of a cpu, so above code is also a window into history.
-- and ++ are special language features that comes nearly for free because of the (virtual) machine architecture.
I produced above from your code example by running IL Disassembly from Microsft .Net.
Kind Regards,
Keld Ølykke
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