A beginner question on how to use a variable as an operator

I have a question could a variable act as an operator such as "+" to add two numbers if so how do i write that? so what I would like to do would be answer = (a b c)
where a is some number b is an operator (such as +) and c is another number, Im not sure how to write that so i could get that line to work and help would be really appreciated

basically trying to a calculator without all the cases

Sorry, but that capability does not exist in C++. You might find it in a dynamically-typed language, although I can't think of one that would allow that.

O_o

Can you explain more why you want the "variable as an operator"?

Soma

What Elkvis said. Not possible in C++, you need a workaround with an if-statement, for example.

Code:

---

if (b == '+') {
    answer = a + b;
} else if (b == '-') {
    answer = a - b;
} else ...

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Quote:

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Originally Posted by spaghetticode

...with an if-statement, for example.

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That's what the OP was hoping to avoid.

Interestingly enough, there's a library available for GCC (C/C++ compiler) called libgccjit, that allows you to generate executable code programmatically, so you could have your calculator take its input, then generate a function based on that input, then call it, and get the result. Much more complex than using switch or if...else if.

Quote:

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Originally Posted by Elkvis

Sorry, but that capability does not exist in C++. You might find it in a dynamically-typed language, although I can't think of one that would allow that.

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You could do it in any language that can interpret its own code as a runtime script (C# could do this, for example, with the Roslyn compiler or CS-Script). Here, your main program would substitute the value of b (i.e. it would create the string "answer = a + c") and then execute that string as code.

That said, I certainly wouldn't use that approach in C#. That would make the solution more complex, not less.

Just for funsies, this is how you do it:

Code:

---

int main()
{
        int x, y;
        std::cout << "Enter two numbers: ";
        std::cin >> x >> y;

        char Op;
        std::cout << "Enter op (+ or -): ";
        std::cin >> Op;
       
        typedef int(__fastcall OpFnc_t)(int, int);
        uint8_t Code[] = { 0x8b, 0xc1, 0x00, 0xc2, 0xc3 };
        uint8_t OpTable['-' + 1];
        OpTable['+'] = 0x03;
        OpTable['-'] = 0x2B;
        Code[2] = OpTable[Op];
        auto * CodeBuf = VirtualAlloc(nullptr, 4096, MEM_COMMIT, PAGE_EXECUTE_READWRITE);
        memcpy(CodeBuf, Code, sizeof(Code) / sizeof(Code[0]));
        OpFnc_t* OpFnc = (OpFnc_t*)CodeBuf;

        int z = OpFnc(x, y);
        std::cout << z << "\n";
}

---

Compiles and runs on VS2015. This goes without saying, but... DON'T DO THIS IN YOUR PRODUCTION CODE. This is just for fun.

https://youtu.be/UL3FHtCq97U

Quote:

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Originally Posted by Elysia

Just for funsies, this is how you do it:

Code:

---

int main()
{
        int x, y;
        std::cout << "Enter two numbers: ";
        std::cin >> x >> y;

        char Op;
        std::cout << "Enter op (+ or -): ";
        std::cin >> Op;
       
        typedef int(__fastcall OpFnc_t)(int, int);
        uint8_t Code[] = { 0x8b, 0xc1, 0x00, 0xc2, 0xc3 };
        uint8_t OpTable['-' + 1];
        OpTable['+'] = 0x03;
        OpTable['-'] = 0x2B;
        Code[2] = OpTable[Op];
        auto * CodeBuf = VirtualAlloc(nullptr, 4096, MEM_COMMIT, PAGE_EXECUTE_READWRITE);
        memcpy(CodeBuf, Code, sizeof(Code) / sizeof(Code[0]));
        OpFnc_t* OpFnc = (OpFnc_t*)CodeBuf;

        int z = OpFnc(x, y);
        std::cout << z << "\n";
}

---

Compiles and runs on VS2015. This goes without saying, but... DON'T DO THIS IN YOUR PRODUCTION CODE. This is just for fun.

https://youtu.be/UL3FHtCq97U

---

Wow. That has to be the jankiest way I've ever seen of embedding assembly (or rather, machine) language instructions. I am equal parts impressed and horrified :D

Quote:

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Originally Posted by DawnOfMe

what I would like to do would be answer = (a b c)
where a is some number b is an operator (such as +) and c is another number

---

You can use function pointers for this. (a b c) is the same as saying b(a,c), where b is a function that performs arithmetic such as adding two numbers. For example, define these two functions:

Code:

---

int op_plus(int a, int b)
{
    return a+b;
}

int op_minus(int a, int b)
{
    return a-b;
}

---

Now the appropriate function can be placed in a function pointer at runtime:

Code:

---

int main()
{
    int a = 100;
    int c = 200;

    cout << "Enter + or -: ";
    char q;
    cin >> q;

    typedef int (*FT)(int,int);
    FT b = nullptr;

    switch(q) {
    case '+': b = static_cast<FT>(op_plus);
        break;
    case '-': b = static_cast<FT>(op_minus);
        break;
    }

    if (b) {
        cout << "answer = " << b(a,c) << endl;
    }
}

---

There are many ways to map your operator input parameter '+', '-', etc into operations. And you'll probably extend and change it as development goes on, so it might be a good idea to move the function pointer stuff into a class:

Code:

---

struct Operation {
    typedef int (*FT)(int,int);
    FT b = nullptr;

    Operation(char q)
    {
        switch(q) {
        case '+': b = static_cast<FT>(op_plus);
            break;
        case '-': b = static_cast<FT>(op_minus);
            break;
        }
    }

    int operator()(int a, int c) const { return b(a,c); }
    operator int() const { return (int)b; }
};

---

Now the main code becomes:

Code:

---

#include <iostream>
using namespace std;

/* ... declarations here */

int main()
{
    int a = 100;
    int c = 200;

    cout << "Enter + or -: ";
    char q;
    cin >> q;

    Operation b(q);

    if (b) {
        cout << "answer = " << b(a,c) << endl;
    }
}

---

Quote:

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Originally Posted by c99tutorial

You can use function pointers for this. (a b c) is the same as saying b(a,c), where b is a function that performs arithmetic such as adding two numbers.

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This crossed my mind the first time I read post #1, but the problem is that DawnOfMe was "basically trying to a calculator without all the cases". With a function pointer or function object approach, you still need to "map your operator input parameter '+', '-', etc into operations", and for that you will most likely consider the various cases, whether you literally use a switch with case or an if-else chain.

Quote:

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Originally Posted by c99tutorial

And you'll probably extend and change it as development goes on, so it might be a good idea to move the function pointer stuff into a class:

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The static_cast to FT is not necessary. Instead of the conversion function to int, define an explicit conversion function to bool:

Code:

---

explicit operator bool() const
{
    return b != nullptr;
}

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Though it might be better to throw an exception from the constructor instead.

Then again, if you are going with a class, maybe you might as well go with a function object class hierarchy.