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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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Okay, just to recap… I'm going to show the example they gave me first… Then tell you how I understand what you guys have been trying to help me with.
Then I will post my exercise towards the end and try to explain how I came up with my answer doing it in my head before running the program and printing it to the console window.
<pre lang="c#">
// Example 3-2.cs
// Increment and Decrement
using System;
class ArithmeticOperators
{
public static void Main()
{
int x = 10;
int y = 100;
int z = y-- + x;
Console.WriteLine(z); // result = 110
Console.WriteLine(y); // result = 99 — The value of y after
// decrementing
z = --z + x;
Console.WriteLine(z); // result = 119
}
}
</pre>
So, we print out the variable z… Like it says, the result is 110. Okay, easy enough.
Just in my mind, the variable y automagically becomes 99.
Anyway, you guys have explained to me that the variable y takes on the new value after evaluating the expression.
Semi-sort of easy enough.
Using this as somewhat of a template for my exercise is probably what threw me off.
In the exercise, I got the first expression done in my head correctly. Meaning it matched what the console window printed out.
The second expression, using the example above.
I also thought the variable y would become the new value after evaluating the first expression.
That's why in my head, I came up with 121… The console window printed out 111.
Drill 3-1
Start with the following assignments:
int x = 10;
int y = 100;
int z = y;
Then write a C# program to compute and display the values of the
variables y and z after executing these expressions:
y = y++ + x;
z = ++z + x;
<a href="http://www.widmarkrob.com">My Online Journey</a>
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WidmarkRob wrote: int z = y-- + x;
WidmarkRob wrote: Anyway, you guys have explained to me that the variable y takes on the new value after evaluating the expression.
y takes on the new value of y, which is sort of tautological..
One is subtracted from y, that is all.
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Yes, I get that now…
Post-fix decrementing…
It takes on the new value after evaluating the expression.
On to my new problem.
I used this same type of thinking with my so-called simple exercise I was given.
y = y++ + x;
z = ++z + x;
In using that same thought process of y taking on the new value after evaluating the expression, in this new problem (the first expression).
I did this in my head before printing it to the console window and came up with 121.
The console window printed 111.
In trying to keep things in semi-sort of uniform fashion.
In my head, I also thought this variable y would take on the new value as well.
Where did I go wrong in my head?
<a href="http://www.widmarkrob.com">My Coding Journey</a>
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int x = 10;
int y = 100;
int z = y;
Then write a C# program to compute and display the values of the
variables y and z after executing these expressions:
y = y++ + x;
z = ++z + x;
This exercise is virtually the same as the example they gave to me in the beginning. (My original question I was asking about)
When I print the variable y to the console window, the result is 110. (Which is what I came up but in my head)
Using the same thought process as in the example, where this variable y also takes on the new value (110)…
I used this new value in my head.
This so-called simple exercise is using "Boxing"? (Just throwing in another term that I learned, hopefully using it correctly)
the variable z now equals the variable y. (Which in my mind, I thought would equal 110)
The second expression.
Prefixes incrementing.
So I thought, 110+1+10.
That's how I came up with 121 in my head.
When I went to go print the variable z to the console window, the result was 111.
<a href="http://www.widmarkrob.com">My Coding Journey</a>
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It's pretty simple:
int x = 10;
int y = 100;
int z = y;
At the end of this, you have three variables, each with separate values:
x y z
10 100 100 Although "y" and "z" contain the same number, they aren't linked together, so changing one does not change the other. Think of them as three pockets: two in your trousers, and one in your shirt. If you put ten coins in each pocket, they all have the same number. If you then take 5 coins out of your shirt pocket, you still have ten coins in each of your trouser pockets.
y = y++ + x; Equates to:
int y2 = y;
y = y + 1;
y = y2 + x;
Which means that "y" ends up holding 110 - as you have seen.
z = ++z + x; Is the equivalent of:
int z2 = z + 1;
z = z2 + x; Which is the same as saying:
z = 11 + 100; Which gives you the answer: 111
The universe is composed of electrons, neutrons, protons and......morons. (ThePhantomUpvoter)
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where it confuses me, if you look at the example I posted… The variable y takes on a new value after evaluating the expression.
In my head, I was applying that same logic/thinking too my exercise.
I also thought, in my exercise… That the variable y would also take on the new value (in this case, 110) after evaluating the first expression.
<a href="http://www.widmarkrob.com">My Coding Journey</a>
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It does. But since you don't use "y" in the second expression that doesn't have any effect. Changing the value in "y" doesn't affect the value in "x" or "z" - it's like they are separate pockets. Adding coins to your shirt pocket doesn't affect the number of coins in either of your trouser pockets!
So:
int x = 10;
int y = 100;
int z = y;
y = y++ + x;
z = ++z + x;
The universe is composed of electrons, neutrons, protons and......morons. (ThePhantomUpvoter)
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Okay, I get what you're saying about the different pockets.
This variable y in this exercise is clearly not the same as the variable y in the example.
In the example: y takes on the new value after evaluating the expression. Okay, I understand (sort of)
In the exercise: y does not take on a new value after evaluating the first expression. Different pockets and something like that. Okay, I guess my question is:
When it is postfix, the variable takes on the new value?
When it is prefix, the variable does not take on the new value?
<a href="http://www.widmarkrob.com">My Coding Journey</a>
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No, no - sorry, I'm probably confusing you. Let's go right back to the beginning.
Assume we have a function with just a two lines of code:
int x = 10;
Console.WriteLine(x);
If we run it, we obviously get "10" printed. So let's start with that and add bits between the two lines.
int x = 10;
x++;
Console.WriteLine(x); Will print 11, because the post increment adds one to the value of x
int x = 10;
++x;
Console.WriteLine(x); Will also print 11, because the pre increment adds one to the value as well.
Prefix and postfix operations always affect the value of the variable; the only difference is when the variable is altered.
With prefix, the variable is altered before it's value is used: So
int x= 10;
int y = ++x;
Console.WriteLine("x={0}, y = {1}", x, y);will give us "x = 11, y = 11" because x was altered before it's value was used in the expression. This is the equivalent of writing:
int x = 10;
x = x + 1;
int y = ++x;
Console.WriteLine("x={0}, y = {1}", x, y);With post fix, the variable is altered after its's value is used in the expression:
int x = 10;
int y = x++;
Console.WriteLine("x={0}, y = {1}", x, y);Will give us "x=11, y = 10" and is the equivalent of writing:
int x= 10;
int y = x;
x = x + 1;
Console.WriteLine("x={0}, y = {1}", x, y);
Does that make sense so far?
The universe is composed of electrons, neutrons, protons and......morons. (ThePhantomUpvoter)
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Yes, it is all a little confusing… While I was reading what you are trying to explain. I was trying to do it in my head before going on to reading the next few lines.
The postfix I ended up getting them backwards in my head.
x=10
and
y=11
you said the variable is altered after its value is used.
I think I'm getting confused on exactly which variable.
x is the variable with the postfix attached to it, so… That is the one I'm thinking that it's altered.
When in fact it is the variable y that is assigned to the variable x that is being altered.
<a href="http://www.widmarkrob.com">My Coding Journey</a>
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