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.
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
Perhaps *(p += 1) will be to your liking?
Much better… but can we make it
*((void*)(p = p + 1))?How about some JavaScript
p+=[]**[]?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.
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?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
p += 1 a = *p, 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
(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.