Incrementing and Decrementing - Just Trying to Understand
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"Console.WriteLine(z); // result = 110" After reading the book, that line right there makes sense to me. It said something similar to, "you evaluate the expression before you do that decrement". I get that part, the way the expression was written I understand that. So when we write the next line to the console, "Console.WriteLine(y);" the variable y automagically turned into 99? y is declared at 100. I just don't see how it could automagically turn into 99 without some sort of loop. y-- is used in an expression, not used as a declared variable. I think that's where I'm getting confused. Maybe I've just seen too many YouTube videos where they are using incrementing and decrementing with loops. My Coding Journey
WidmarkRob wrote:
the variable y automagically turned into 99?
But it is not automagically turned in to 99 that is what the decrement operator does! And that is what I have been telling you. Take your code:
int z = y-- + x;
Substitute your values
int z = 100-- + 10;
The value of z is calculate right?
100 + 10 = 110
right? The then post decrement operator is applied to y so
y = y - 1
which make
y = 99
. Still not sure why you would think that you would need a loop to add or subtract anything.
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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? My Coding Journey
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
xis 2, because it was incremented twice. Andzis also 2, proving that the increment really happens while the expression is evaluated. Firstx++is evaluated, the result is 0 and the effect is thatxis now 1. Then++xis evaluated,xwas 1 and for pre-increment the result is the value after the increment, so the result is two (the effect is again thatxis incremented, of course). So that works out to0 + 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,
xis 2, obviously, andarraylooks like this:{ 1, 0 }What happened, was the firstx++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 thenxis incremented again, and finally the value of the right hand side (which was 1) is stored in the array at index 0. -
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
xis 2, because it was incremented twice. Andzis also 2, proving that the increment really happens while the expression is evaluated. Firstx++is evaluated, the result is 0 and the effect is thatxis now 1. Then++xis evaluated,xwas 1 and for pre-increment the result is the value after the increment, so the result is two (the effect is again thatxis incremented, of course). So that works out to0 + 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,
xis 2, obviously, andarraylooks like this:{ 1, 0 }What happened, was the firstx++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 thenxis incremented again, and finally the value of the right hand side (which was 1) is stored in the array at index 0.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. :-) My Coding Journey
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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
xis 2, because it was incremented twice. Andzis also 2, proving that the increment really happens while the expression is evaluated. Firstx++is evaluated, the result is 0 and the effect is thatxis now 1. Then++xis evaluated,xwas 1 and for pre-increment the result is the value after the increment, so the result is two (the effect is again thatxis incremented, of course). So that works out to0 + 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,
xis 2, obviously, andarraylooks like this:{ 1, 0 }What happened, was the firstx++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 thenxis incremented again, and finally the value of the right hand side (which was 1) is stored in the array at index 0.:confused: 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 My Coding Journey
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:confused: 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 My Coding Journey
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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:confused: 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 My Coding Journey
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
forloops.The universe is composed of electrons, neutrons, protons and......morons. (ThePhantomUpvoter)
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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
forloops.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.
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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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..
Do you want to sit there and work out why
y++ + ++y * y--equals 154? Quantum magic works for me! :laugh:The universe is composed of electrons, neutrons, protons and......morons. (ThePhantomUpvoter)
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Do you want to sit there and work out why
y++ + ++y * y--equals 154? Quantum magic works for me! :laugh: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.
:laugh: 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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WidmarkRob wrote:
It is declared at 100.
Yes the initial value is 100, did you see the y--? As you were told, the -- is the decrement operator that makes y 99. The -- is short hand for y = y - 1.
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.A guide to posting questions on CodeProject[^]
Dave Kreskowiak -
:laugh: 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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Now you see why I hit people who do it! :laugh:
The universe is composed of electrons, neutrons, protons and......morons. (ThePhantomUpvoter)
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Now you see why I hit people who do it! :laugh:
The universe is composed of electrons, neutrons, protons and......morons. (ThePhantomUpvoter)
.. And if they turn the other cheek, hit them again :laugh: This has got to be close to using GoTo statments or possibly worse....
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.. And if they turn the other cheek, hit them again :laugh: This has got to be close to using GoTo statments or possibly worse....
I think of it as "Codefuscating" ;)
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 :laugh: 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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Worse. With goto at least only one thing can happen, even if it's not always immediately clear what that thing is.
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.
