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Home/ Questions/Q 8449597
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Editorial Team
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Editorial Team
Asked: June 10, 20262026-06-10T10:45:13+00:00 2026-06-10T10:45:13+00:00

I’ve been back & forth with this problem for a while especially since I

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I’ve been back & forth with this problem for a while especially since I started to OpenCV library. The fact is, in OpenCV, there are several methods used:

  • 1st: funcA((const) CvMat arg)
  • 2nd: funcA((const) CvMat& arg)
  • 3rd: funcA((const) CvMat* arg)
  • 4th: funcA((const) CvMat*& arg) => I’ve just seen and currently been stuck at this

and of course, corresponding to each method, the caller format and the function implementation should be different.

What is the significance about all of these derivatives?? especially the last one (I’ve not yet understood its usage)

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  1. Editorial Team
    Editorial Team
    2026-06-10T10:45:14+00:00Added an answer on June 10, 2026 at 10:45 am

    Ignoring the (const) for now, and using int for clarity:

    Pass by value makes a copy in the body of the function

    void funcA(int arg) {
      // arg here is a copy
      // anything I do to arg has no effect on caller side.
      arg++; // only has effect locally
    }
    

    Note that it semantically makes a copy, but the compiler is allowed to elide the copies under certain conditions. Look up copy elision

    Pass by reference. I can modify the argument passed by the caller.

    void funcA(int& arg) {
      // arg here is a reference
      // anything I do to arg is seen on caller side.
      arg++;
    }
    

    Pass pointer by value. I get a copy of the pointer, but it points to the same object pointed at by the caller’s argument

    void funcA(int* arg) {
      // changes to arg do not affect caller's argument
      // BUT I can change the object pointed to
      (*arg)++; // pointer unchanged, pointee changed. Caller sees it.
    }
    

    Pass reference to pointer. I can change the pointed itself and the caller will see the change.

    void funcA(int*& arg) {
      // changes to arg affect caller's argument
      // AND I can change the object pointed to.
      (*arg)++; // pointee changed
      arg++; // pointer changed. Caller sees it.
    }
    

    As you can see, the second two are just the same as the first two, except that they deal with pointers. If you understand what pointers do, then there is no difference conceptually.

    Concerning const, it specifies whether the argument can be modified or not, or, if the arguments are references or pointers, whether what they point to/refer to can be modified. The positioning of const is important here. See const correctness for example.

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