How to pass a struct member pointer to a function?

Question

How does one go about passing a struct member identifier to a function such that the function can apply an operation to that specified member? The below code example should highlight what I am trying to achieve.

struct _struct {
    int a = 0;
    int b = 0;
};
_struct test[5];

void printVar(int index, ??? member) {
    // Print specified member at array element
    printf(test[index].member);
};

int operation(??? member) {
    // Some operation applied to specified member of all array elements
    // eg. Averaging all the readings contained in test[].a
    int total = 0;
    for (int i = 0; i < 5; i++) {
        total += test.member;
    };
    return total / 5;
};

The closest I have found to a solution is described at the below link, whilst the principle is sound in its application it doesn't appear to be the best possible implementation. http://cplusplus.com/forum/beginner/45268/

You could use a std::map or std::unordered_map instead of a struct, and then pass the desired key as the member identifier. Alternatively, you could pass a std::function that fetches the desired member from the an argument of type _struct. — Bungo
– Bungo, Commented Feb 20, 2021 at 12:01
If I understand it correctly, you want to have global array test and pass either member a or b to operation, so that it returns sum of all a or b from the array? — Yksisarvinen
– Yksisarvinen, Commented Feb 20, 2021 at 12:13

PeterT · Accepted Answer · 2021-02-20 12:20:51Z

3

This is what "pointer to member" is for.

struct _struct {
    int a = 0;
    int b = 0;
};
_struct test[5];

void printVar(int index, int _struct::* member) {
    // Print specified member at array element
    printf("%d\n",test[index].*member);
};

int operation(int _struct::* member) {
    // Some operation applied to specified member of all array elements
    // eg. Averaging all the readings contained in test[].a
    int total = 0;
    for (int i = 0; i < 5; i++) {
        total += test[i].*member;
    };
    return total / 5;
};


int main(){
    printVar(1,&_struct::b);
    operation(&_struct::a);
    return 0;
}

answered Feb 20, 2021 at 12:20

PeterT

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3 Comments

Branyon Apel Over a year ago

That is exactly the solution I needed, thankyou! Funny how no amount of googling reveals what you want sometimes. I had actually even tried what you suggested minus the "int" datatype of the pointer.

Branyon Apel Over a year ago

@Scheff Would have done so earlier if the option had been there, I upvoted instead.

Scheff's Cat Over a year ago

@BranyonApel This might be caused by your rep. Once, I made a question and got an answer immediately, I had to wait 10 minutes but my rep. was already higher... ;-)

Bungo · Accepted Answer · 2021-02-20 12:23:22Z

2

How about passing a std::function that returns the desired member from a given _struct? Like this:

#include <functional> // for std::function

struct _struct {
    int a = 0;
    int b = 0;
};
_struct test[5];

void printVar(int index, std::function<int(const _struct&)> getMember) {
    // Print specified member at array element
    printf("%d\n", getMember(test[index])); // note: added missing "%d"
};

int operation(std::function<int(const _struct&)> getMember) {
    // Some operation applied to specified member of all array elements
    // eg. Averaging all the readings contained in test[].a
    int total = 0;
    for (int i = 0; i < 5; i++) {
        total += getMember(test[i]); // note: added missing [i]
    };
    return total / 5;
};

You would use it like this:

int get_a(const _struct& s) { return s.a; }
int get_b(const _struct& s) { return s.b; }

int main()
{
    test[0] = {0,1};
    test[1] = {2,3};
    test[2] = {4,5};
    test[3] = {6,7};
    test[4] = {8,9};

    printVar(2, get_a); // prints value of test[2].a
    printVar(4, get_b); // prints value of test[4].b

    const int result_a = operation(get_a); // perform operation on the a members
    const int result_b = operation(get_b); // perform operation on the b members

    return 0;
}

Or, if you're familiar with lambdas, you can use them to define get_a and get_b instead of making them free-standing functions:

int main()
{
    test[0] = {0,1};
    test[1] = {2,3};
    test[2] = {4,5};
    test[3] = {6,7};
    test[4] = {8,9};

    const auto get_a = [](const _struct& s) { return s.a; };
    const auto get_b = [](const _struct& s) { return s.b; };

    printVar(2, get_a); // prints value of test[2].a
    printVar(4, get_b); // prints value of test[4].b

    const int result_a = operation(get_a); // perform operation on the a members
    const int result_b = operation(get_b); // perform operation on the b members

    return 0;
}

edited Feb 20, 2021 at 12:23

answered Feb 20, 2021 at 12:16

Bungo

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3 Comments

Scheff's Cat Over a year ago

You even could use a lambda instead of get_a() and get_b(), e.g.: const int result_a = operation([](const _struct &s) { return s.a; });

Bungo Over a year ago

@Scheff Yep, I was going to use a lambda but wasn't sure if the OP was familiar with them. Updated to add a version that defines them as lambdas.

Branyon Apel Over a year ago

Hi, thanks for your answer. Thats an interesting way to do it, however @PeterT has exactly the methodology I was chasing. Having to define the separate function for each new parameter you have to grab adds the extra step as such.

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