CODEWITHSUNDEEP
c++expressionconstants
Tivi
Tivi

Reputation: 496

C++ what should I use instead of #define

I would like to give a value from the console with cin operator instead of using a #define N 6. I've tried it but I get "expression must have constant value" error message. How should I do it by the other way?

Thank you for answers!

Example code:

#include <iostream>
#include <string>
#include <cstdlib> 
#include <ctime>
#define N 6
using namespace std;

typedef struct person {
    int roll;
    string name;
} Person;

int main() {
    int numberofperson;
    cout << "Number of people: "; cin >> numberofperson;
    srand(time(NULL));
    Person people[N];
    int i;
    for (i = 0; i < numberofperson; i++)    {
        cout << "Write the " << i + 1 << ". name of the person: ";
        cin  >> people[i].name;
        people[i].roll = rand() % 6 + 1;
        cout << "Roll with dice: " << people[i].roll<<endl;
    }

    return 0;
}

Upvotes: 4

Views: 349

Answers (1)

Ivan Smirnov
Ivan Smirnov

Reputation: 4435

Defines are unrolled with the preprocessor. In fact, #define N 6 means that all occurrences of N in your code will be replaced by 6, thus replacing cin >> N with cin >> 6.

The solution is to make N a variable:

int N;
cin >> N;
// do whatever you want with N

However, note that in this case Person people[N] is a variable-sized array (that is, its size is not known during compilation time. It is non-standard C++ and you should avoid it. Consider using vector instead – it is basically a variable-sized array from the standard library.

cin >> N;
vector<Person> people(N);
...
cin >> people[i].name;

Upvotes: 5

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