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k1k4ss0
Asked: 2020-02-29 07:40:56 +0800 CST 2020-02-29 07:40:56 +0800 CST

error with templates and method

  • 772

Hello, I am making a code with templates, and when I compile it I get the following error:

    main.cpp: In function ‘int main()’:
    main.cpp:41:28: error: ‘class number<char, 0, 10>’ has no member named ‘make_op’
             case '+': number_T.make_op(operacion);break;
                                ^~~~~~~
    main.cpp:42:28: error: ‘class number<char, 0, 10>’ has no member named ‘make_op’
             case '-': number_T.make_op(operacion);break;
                                ^~~~~~~
    main.cpp:43:28: error: ‘class number<char, 0, 10>’ has no member named ‘make_op’
             case '*': number_T.make_op(operacion);break;
                                ^~~~~~~
    main.cpp:44:28: error: ‘class number<char, 0, 10>’ has no member named ‘make_op’
             case '/': number_T.make_op(operacion);break;
                                ^~~~~~~
    main.cpp:45:28: error: ‘class number<char, 0, 10>’ has no member named ‘make_op’
             case '=': number_T.make_op(operacion);break;
                                ^~~~~~~
    main.cpp:46:28: error: ‘class number<char, 0, 10>’ has no member named ‘make_op’
             case '<': number_T.make_op(operacion);break;
                                ^~~~~~~
    main.cpp:47:28: error: ‘class number<char, 0, 10>’ has no member named ‘make_op’
             case '>': number_T.make_op(operacion);break;
                                ^~~~~~~

The code in the main that you mention is the following:

    switch(operacion){                                      // ejecutamos las operaciones

        case '+': number_T.make_op(operacion);break;
        case '-': number_T.make_op(operacion);break;
        case '*': number_T.make_op(operacion);break;
        case '/': number_T.make_op(operacion);break;
        case '=': number_T.make_op(operacion);break;
        case '<': number_T.make_op(operacion);break;
        case '>': number_T.make_op(operacion);break;
    }

And the file that contains the object and the method is the following:

#include <iostream>
#include <cmath>
using namespace std;
template < class T, size_t N, size_t B >

class number{

    private:
    T* digitos; 
    int index;
    int base;
    int val1,val2;
    int final=0;
    int size;
    public:



    void s_base(int i){
        base=i;

    }       
    void set_vals(int a, int b){
        val1=a;
        val2=b;

    }


    number(T valor):digitos(NULL){
       resize_v(N);
       int sz = sizeof(valor);
       resize_v(sz);       
    }


    void resize_v(int sz){

        digitos = new T [sz];
        size=sz;
        index = sz -1;
    }


    int longitud(int number) const{

            return ceil(log(number)/log(base));  // usa la libreria math.h

    }



    void set_base(int n ){
        if(n>=base){
            int result = n / base;             // obtenemos el valor resultado de la divicion, para la llamada recursiva
            int rest = n % base;               // obtenemos el resto, para guardarlo en el vector
            if(rest > 9 ){                  // en caso que el numero dado sea mayor que 9, hay que representarlo en caracteres alfabeticos
                rest+=55;                    // si rest = 10, 10 + 55 = 65 === 'A'
                digitos[index]=(char)rest;
            }
            else{
                rest+=48;
                digitos[index]=(char)rest;
            }
            index--;                        // disminuimos el indice del vector
            set_base(result);               // llamada recursiva
        }


        else{
            if(n > 9 ){                  // en caso que el numero dado sea mayor que 9, hay que representarlo en caracteres alfabeticos
                n+=55;                    // si rest = 10, 10 + 55 = 65 === 'A'
                digitos[index]=(char)n;
            }
            else{
                n+=48;
                digitos[index]=(char)n;
            }
            digitos[index]=n;                 // al llegar al ultimo valor en el que no se pueda dividir por la base, se inserta en la posicion mas
            index=0;                        // significativa
        }
    }

    void make_op(char op){

        int result=0;
        switch(op){

        case '+':  result=val1+val2;resize_v(longitud(result));set_base(result);break;
        case '-':  result=val1-val2;resize_v(longitud(result));set_base(result);break;
        case '*':  result=val1*val2;resize_v(longitud(result));set_base(result);break;
        case '/':  result=val1/val2;resize_v(longitud(result));set_base(result);break;
        case '=':  val1==val2 ? cout<<"Son iguales"<<endl : cout<<"No son iguales"<<endl;break;
        case '<':  val1<val2 ? cout<<"El primer valor es menor"<<endl : cout<<"El primer valor es mayor";break;
        case '>':  val1>val2 ? cout<<"El primer valor es mayor"<<endl : cout<<"El primer valor es menor"<<endl;break;

        }

    }
    int get_result(){
        if(index>0){

            int result=int(digitos[index]);
            result>=55 ? result-55 : result-48;
            int a=pow(base,index);
            index--;

            final=get_result()+ result*a;

            return final;
        }

    }


    ostream& write(ostream& os)const{

        for(int i = 0; i <= size; i++){
            cout << " | " << digitos[i];
        }
        cout<< endl;

    }   
};
c++

1 Answers

  1. Best Answer

    That template leaks everywhere:

    Builder

    The template's constructor is supposed to receive a value... but it doesn't do anything with that value. I would expect it to store it in digitos, but no.

    number(T valor):digitos(NULL){
   resize_v(N);
   int sz = sizeof(valor);
   resize_v(sz);       
}

    The only thing it does with valoris call sizeoffor what? that operation does not make sense since it sizeofwill not tell you how many digits the number has. sizeofis evaluated at compile time and tells you how many bytes the type in question occupies.

    • Yes T == int, sizeof(valor) == 4and it doesn't matter the content ofvalor
    • If T == char[],sizeof(value) == strlen(value)+1`. You also don't get the number of digits
    • yes T == char,sizeof(valor) == 1
    • ...

    Taking into account that Tit is used as the base type of the array, we see that it would only make sense that T == char, but in that case we can only put a single digit in the object... it doesn't make sense.

    memory leaks

    Every time you call you resize_vcreate a memory leak. you don't free the memory before calling new:

    void resize_v(int sz){

    digitos = new T [sz]; // <<--- que pasa con la reserva anterior?
    size=sz;
    index = sz -1;
}

    you need return

    If a function has a return type, it must necessarily have a return . In your case that is not fulfilled in the function write:

    ostream& write(ostream& os)const{

    for(int i = 0; i <= size; i++){
        cout << " | " << digitos[i];
    }
    cout<< endl;
                              // <<--- Aqui falta un return
}

    unused types

    The template has 3 parameters. What does each one work for?

    Tis used as the base for the array digitos. Does it make sense for it to digitosbe an array of double? it seems to me not. It is also used in the constructor, here it could make sense but only to be able to pass from Tto an array of integers or characters. From my point of view Tit should not be a parameter of the class but of the constructor:

    template<class T>
number(T valor)
{
   // ...
}

    Nis used to determine the size of the array digitos, but only temporarily because it is then resized to a different size. What is it for Nthen?

    BIt has no use directly, so it is a parameter that seems to be redundant.

    In short, with the code you present there is no reason for it to numberbe a template.

    class number{
private:
    char* digitos; 
    int index;
    int base;
    int val1,val2;
    int final=0;
    int size;

public:
    template<class T>
    number(T valor)
    {
        // ...
    }
};

    But of course, I don't quite understand what utility this class intends to cover. You may need to rephrase your question to better refine the answers.

    • 1