In the screenshot it said x = *(++p) and iirc that is not the same as saying x = *(p++) or x = *(p += 1)
As in my example using ++p will return the new value after increment and p++ or p+=1 will return the value before the increment happens, and then increment the variable.
Or at least that is how I remember it working based on other languages.
I’m not sure what the * does, but I’m assuming it might be a pointer reference? I’ve never really learned how to code in c or c++ specifically. Though in other languages ( like PHP which is based on C ) there is a distinct difference between ++p and (p++ or p+= 1)
The last two behave the same. Though it has been years since I did a lot of coding. Which is why I asked.
I’ll install the latest PHP runtime tonight and give it a try xD
But assuming it doesn’t the context is p_ch = the bits above… the code declaring p_ch isn’t shown but I’m guessing that the value here is actuality a pointer to a pointer so nothing illegal would be happening.
Lastly… C++ is really lacking in guarantees so you can assign a char to the first byte of an integer - C++ doesn’t generally care what you do unless you go out of bounds.
The reason I’m casting to void* is just pure comedy.
That’s not a real operator. You’ve put a space in “i–” and removed the space in “-- >”. The statement is “while i-- is greater than zero”. Inventing an unnecessary “goes to” operator just confuses beginners and adds something else to think about while debugging.
And yes I have seen beginners try to use <-- and --<. Just stop it.
The sheer number of people that do not expect a joke on this community… (Really, if you are trying to learn how to program pay attention to the one without the Humor on the name, not here.)
It requires some decent knowledge of dereferencing and C++ “++” operator overloads. It’s not exactly trivial to glance at and feel confident you know what’s happening
I didn’t know why, but *++p bugs me
Perhaps *(p += 1) will be to your liking?
p = 1 x = ++p // x = 2 // p = 2p = 1 x = p++ // x = 1 // p = 2++pwill increase the value and return the new valuep++will increase the value and return the old valueI think
p = p + 1is the same asp++and not as++p. No?(p += 1) resolves to the value of p after the incrementation, as does ( p = p + 1).
Yes.
p++==p+= 1==p = p + 1are all the same if you use it in an assignment.++pis different if you use it in an assignment. If it’s in its own line it won’t make much difference.That’s the point I was trying to make.
No.
++p returns incremented p.
p += 1 returns incremented p.
p = p + 1 returns incremented p.
p++ returns p before it is incremented.
Right. So i had them the other way around. :D
Thanks for clarifying.
In C an assignment is an expression where the value is the new value of what was being assigned to.
In
a = b = 1, both a and b will be 1.a = *(p = p + 1)is the same as
, so ++p.
What I meant was:
In the screenshot it said
x = *(++p)and iirc that is not the same as sayingx = *(p++)orx = *(p += 1)As in my example using ++p will return the new value after increment and p++ or p+=1 will return the value before the increment happens, and then increment the variable.
Or at least that is how I remember it working based on other languages.
I’m not sure what the * does, but I’m assuming it might be a pointer reference? I’ve never really learned how to code in c or c++ specifically. Though in other languages ( like PHP which is based on C ) there is a distinct difference between
++pand (p++orp+= 1)The last two behave the same. Though it has been years since I did a lot of coding. Which is why I asked.
I’ll install the latest PHP runtime tonight and give it a try xD
Much better… but can we make it
*((void*)(p = p + 1))?Why are you casting to
void*? How is the compiler supposed to know the size of the data you are dereferencing?This would probably cause a compiler error…
But assuming it doesn’t the context is
p_ch =the bits above… the code declaring p_ch isn’t shown but I’m guessing that the value here is actuality a pointer to a pointer so nothing illegal would be happening.Lastly… C++ is really lacking in guarantees so you can assign a char to the first byte of an integer - C++ doesn’t generally care what you do unless you go out of bounds.
The reason I’m casting to void* is just pure comedy.
How about some JavaScript
p+=[]**[]?That
*++operator from C is indeed confusing.Reminds me of the goes-to operator:
-->that you can use as:That’s not a real operator. You’ve put a space in “i–” and removed the space in “-- >”. The statement is “while i-- is greater than zero”. Inventing an unnecessary “goes to” operator just confuses beginners and adds something else to think about while debugging.
And yes I have seen beginners try to use <-- and --<. Just stop it.
The sheer number of people that do not expect a joke on this community… (Really, if you are trying to learn how to program pay attention to the one without the
Humoron the name, not here.)Well, I guess nobody expects.
It requires some decent knowledge of dereferencing and C++ “++” operator overloads. It’s not exactly trivial to glance at and feel confident you know what’s happening
I thought it was just incrementing the address and dereferencing it, but I don’t write C or C++. What is being overloaded there?
It’s very likely plain old C
Oh yeah Linus famously hates C++. Still confusing syntax imo
Not if you’ve done a lot of pointer math
welcome to C