• @[email protected]
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          38 months ago

          Why are you casting to void*? How is the compiler supposed to know the size of the data you are dereferencing?

          • @[email protected]
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            8 months ago

            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.

      • @fluckx
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        8 months ago
        p = 1
        
        x = ++p
        // x = 2
        // p = 2
        
        p = 1
        x  = p++
        // x = 1
        // p = 2
        

        ++p will increase the value and return the new value

        p++ will increase the value and return the old value

        I think p = p + 1 is the same as p++ and not as ++p. No?

        • @[email protected]
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          8 months ago

          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.

          • @fluckx
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            18 months ago

            What I meant was:

            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

        • @SpaceNoodle
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          8 months ago

          (p += 1) resolves to the value of p after the incrementation, as does ( p = p + 1).

          • @fluckx
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            28 months ago

            Yes.

            p++ == p+= 1 == p = p + 1 are all the same if you use it in an assignment.

            ++p is 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.

            • @SpaceNoodle
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              58 months ago

              No.

              ++p returns incremented p.

              p += 1 returns incremented p.

              p = p + 1 returns incremented p.

              p++ returns p before it is incremented.

              • @fluckx
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                38 months ago

                Right. So i had them the other way around. :D

                Thanks for clarifying.

    • @marcos
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      That *++ operator from C is indeed confusing.

      Reminds me of the goes-to operator: --> that you can use as:

      while(i --> 0) {
      
      • @[email protected]
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        158 months ago

        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.

        • @marcos
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          18 months ago

          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.)

          Well, I guess nobody expects.

      • @allan
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        8 months ago

        It’s very likely plain old C

      • Eager Eagle
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        58 months ago

        I thought it was just incrementing the address and dereferencing it, but I don’t write C or C++. What is being overloaded there?