CODEWITHSUNDEEP
c++arraysfunction
KGS
KGS

Reputation: 685

passing arrays (&) to functions c++, runtime error

#include <iostream>
#include <cmath>
using namespace std;

void input(int(&x)[10]);
int copy(int (&x)[10], int(&y)[10]);
void read(int(&x)[10], int b);


int main()
{
    int m[10], v[10], c; 
    input((&m)[10]);
    read((&m)[10], 10);
    c = copy((&m)[10], (&v)[10]);
    read((&v)[c],c);
    system("PAUSE");
    return 0;
}

void input(int(&x)[10])
{
    for(int i=0; i < 10; i++)
    {
        cout<<"enter number "<<i+1<<": ";
        cin>>x[i];
    }
}

void read(int(&x)[10], int b)
{
    for(int i=0; i < b; i++)
    {
        cout<<"number("<<i+1<<"): "<<x[i];
    }
}

int copy(int(&x)[10], int(&y)[10])
{
    int c = 0;
    for(int i = 0; i < 10; i++)
    {
        if(x[i] % 3 == 0)
        {
            y[c]=x[i];
            c++;
        }
    }
    return c;
}

I get a runtime error when i input the first number, any help will be appreciated I think the problem is in the way i pass the arrays to the functions but I am not really sure the aim of the program is to get input for 10 int-s in the array m, than transfer those divisible by 3 to v, and output both

Upvotes: 1

Views: 870

Answers (6)

Cabes234
Cabes234

Reputation: 3

It does not work to pass arrays into c++. This will result in a run time error unless you use pointers or create a class that contains an array and pass that.

Upvotes: 0

Lightness Races in Orbit
Lightness Races in Orbit

Reputation: 385274

void input(int(&x)[10])

In this function signature:

  • x is the argument name
  • int(&)[10] is the argument type (reference to an array of 10 ints).

Now consider how you've called this function, presumably by copy/paste:

input((&m)[10]);

With any function call, you write just the name of a variable, not its type; so, when you're passing the array to the function, you should just write:

input(m);

(At present you're passing something else that does not exist in memory, because of &'s triple meaning.)

Same goes for the other places where you pass [references to] arrays.

Upvotes: 0

Some programmer dude
Some programmer dude

Reputation: 409364

You are passing pointers to the functions, not references.

input((&m)[10]);

In the line above, the & is getting the address of m, and making it into a pointer. The functions are not expecting pointers. Skip the & in the calls, and it should work.

Also, I suggest you try to find a tutorial describing the difference between a pointer and a reference. A quick Google search turned up this question here on Stack Overflow, with some good answers.

Upvotes: 0

iammilind
iammilind

Reputation: 70030

The syntax of passing an array is not correct (though it compiles). It should be as simple as,

input(m);
read(m, 10);
c = copy(m, v);
read(v, c);

int(&x)[10] should be used for the function prototype where you are receiving an array of size 10 by reference. While calling the function simply pass the name of the variable.

Upvotes: 3

M. Wangel
M. Wangel

Reputation: 101

Expanding on the answer above: m[10] is past the last element in the array so what happens is you are reading memory outside your variables which is generally considered to be bad form.

Upvotes: -1

Mike Seymour
Mike Seymour

Reputation: 254631

input((&m)[10]);

This doesn't do what you think it does; &m is a pointer to an array, so (&m)[10] is a reference to the 10th element of an array of arrays; in other words, to an invalid block of memory some distance off the end of m.

You want input(m); to pass a reference to m.

Upvotes: 4

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