How to call struct from separate function?

Question

#include <iostream>
#include <cmath>

using namespace std;

struct workers{
    int ID;
    string name;
    string lastname;
    int date;
};

bool check_ID(workers *people, workers &guy);
void check_something(workers *people, workers &guy, int& i);

int main()
{
    workers people[5];
    for(int i = 0; i < 5; i++){
        cin >> people[i].ID;
        cin >> people[i].name;
        cin >> people[i].lastname;
        cin >> people[i].date;
        if(check_ID(people, people[i]) == true)
            cout << "True" << endl;
        else
            cout << "False" << endl;
        check_something(people, people[i], i);
    }
    return 0;
}

bool check_ID(workers *people, workers &guy){
    for(int i = 0; i < 5; i++){
        if(people[i].ID == guy.ID)
            return true;
            break;
    }
    return false;
}

void check_something(workers *people, workers &guy, int& i){
    check_ID(people, guy[i]);
}

This is the code I have, it's not very good example, but I quickly wrote it to represent my problem I have because my project is kinda too big. So basically, I want to call struct from a different function and I'm getting this error: error: no match for 'operator[]' in guy[i] on this line : check_ID(people, guy[i]); in the function check_something.

what's the purpose of guy[i]? if you want to be able to apply subscript operator to struct workers, of course you need to define it, the compiler's making it very clear. — Einheri
– Einheri, Commented Jan 14, 2014 at 3:27
My fault.. I forgot void returns nothing, it seems to work now! :-) Thanks a lot for your help. — user3182683
– user3182683, Commented Jan 14, 2014 at 3:35

kfsone · Accepted Answer · 2014-01-14 03:32:01Z

5

In main, people is an array. You access the ith element of it people[i] and try to pass it to check_something in the position of function-local variable guy. You then try to dereference guy - which is not an array, but a single object instance.

int main()
{
    workers people[5];  // <-- array

...

    check_something(people /* <-- people */, people[i] /* <-- guy */, i /* <-- i */);

vs

void check_something(workers *people, workers &guy, int& i){
    check_ID(people, guy[i] /* <-- array access on single instance*/);

You actually passed the array in the first argument, people. You don't need "guy" here, because it is people[i], isn't it? So you could do:

void check_something(workers *people, int& i){
    worker& guy = people[i];
    check_ID(people, guy);

or just

void check_something(workers *people, int& i){
    check_ID(people, people[i]);

or

would work, or you could just pass

void check_something(workers* people, workers& guy) {
    check_id(people, guy);
}

---- EDIT ----

You also have a python-like bug in your check_ID function.

   if(people[i].ID == guy.ID)
        return true;
        break;

In Python, this says:

if people[i].ID == guy.ID:
    return True

break

What you want is

if ( people[i].ID == guy.ID ) {
    return true;
    break;
}

or just

if ( people[i].ID == guy.ID )
    return true;

(since the return is going to exit the function, there's no point in also saying break afterwards)

edited Jan 14, 2014 at 3:32

answered Jan 14, 2014 at 3:25

kfsone

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5 Comments

user3182683 Over a year ago

I declared workers guy = people[i]; and when I call the function with check_ID(people, guy), It still does nothing (doesn't do its function). Why is that? I also tried people[i] as second reference and it still doesn't output anything.

kfsone Over a year ago

workers guy = people[i] makes a copy of people[i] into a temporary, local variable called guy.

kfsone Over a year ago

Also, your 'check_ID' function is broken because you have a 'break' that always executes.

kfsone Over a year ago

Updated my answer to include the check_ID/return/break thing.

user3182683 Over a year ago

Thanks a lot for the help! I really should start using {} everywhere, even if it's one line code I guess.

yizzlez · Accepted Answer · 2014-01-14 03:24:39Z

0

workers does not have an overloaded subscript operator, nor is guy an array. Therefore you cannot call [] on it. Either create an array or delete [i] after guy.

check_ID(people, guy); //delete [i]

answered Jan 14, 2014 at 3:24

yizzlez

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1 Comment

user3182683 Over a year ago

I did try with guy, but it does nothing.. it goes through the function, does the call to check_ID, but the check_ID does nothing, I guess because the guy has no value or something?

tillaert · Accepted Answer · 2014-01-14 03:26:02Z

0

Guy is a reference, not a pointer. You cannot use the operator[] on a reference.

answered Jan 14, 2014 at 3:26

tillaert

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