How to print data from a structure in a loop - c++

Question

I want a loop that prints the values for each soda.
When I try to print machine[j].name, an error is thrown stating that no operator is found or no acceptable conversion exists.
struct Soda
{
string name;
double price;
int quantity;
} a, b, c, d, e;
int main(int argc, char** argv)
{
Soda machine[5];
a.name = "Cola";
a.price = .80;
a.quantity = 20;
b.name = "Root Beer";
b.price = .75;
b.quantity = 20;
c.name = "Lemon-Lime";
c.price = .90;
c.quantity = 20;
d.name = "Grape - Mango";
d.price = .80;
d.quantity = 20;
e.name = "Cream";
e.price = .80;
e.quantity = 20;
cout << setw(5) << " Name" << setw(15) << " Cost"
<< setw(15) << " Stock" << setw(14) << "";
for (int i = 0; i < 5; i++)
for (int j = 0; j < 3; j++){
cout << left << setw(5) << machine[j].name
<< setw(15) << " Cost" << setw(15) << " Stock"
<< setw(14);
return 0;
}

Answer 1

It appears that you are mixing up the array with the declaration for separate variables. You have set values on the non-array variables and then you are trying to loop through the array.
Instead, you should set the values on the actual elements of the array.
#include <iostream>
#include <string>
using std::string;
using std::cout;
using std::endl;
struct Soda
{
string name;
double price;
int quantity;
};
int main(){
Soda machine[5];
machine[0].name = "Cola";
machine[0].price = .80;
machine[0].quantity = 20;
machine[1].name = "Root Beer";
machine[1].price = .75;
machine[1].quantity = 20;
machine[2].name = "Lemon-Lime";
machine[2].price = .90;
machine[2].quantity = 20;
machine[3].name = "Grape - Mango";
machine[3].price = .80;
machine[3].quantity = 20;
machine[4].name = "Cream";
machine[4].price = .80;
machine[4].quantity = 20;
cout << "Name Cost Stock" << endl;
for (int i = 0; i < 5; i++)
cout << machine[i].name << " " << machine[i].price << " " << machine[i].quantity << endl;
}

Answer 2

You are trying to print out the contents of objects (array of objects ) machine[i], when instead you have assigned values to some different objects: a, b, c, d, e.
Solution:
1. Either print a.name, b.name etc.
2. or instead of
a.name = "Cola";
a.price = .80;
write:
machine[i].name = "Cola";
machine[i].price = .80;

Answer 3

Since you marked this as C++, I suggest you try a vector (for machine).
int main()
{
vector<Soda> machine;
Soda tmpSoda;
{
tmpSoda.name = "Cola";
tmpSoda.price = .80;
tmpSoda.quantity = 20;
machine.push_back(tmpSoda);
}
{
tmpSoda.name = "Root Beer";
tmpSoda.price = .75;
tmpSoda.quantity = 20;
machine.push_back(tmpSoda);
}
// ....
{
tmpSoda.name = "Cream";
tmpSoda.price = .80;
tmpSoda.quantity = 20;
machine.push_back(tmpSoda);
}
Now all soda's are in the vector (named machine), so you can now report each element in a loop.
cout << "Name Cost Stock" << endl;
for (int i=0; i<machine.size(); ++i)
{
cout << machine[i].name << " "
<< machine[i].price << " "
<< machine[i].quantity << endl;
}
Since you marked this as C++, I suggest you try a vector of class instances. Note how the ctor fills in the values of each Soda, and the "show" method uses the private data attributes of each instance.
class Soda
{
public:
Soda(std::string aName,
double aPrice,
int aQuantity) :
name(aName),
price(aPrice),
quantity(aQuantity)
{
}
~Soda() {
name.erase();
quantity = 0;
}
std::string show() {
std::stringstream ss;
ss << std::right << std::setw(14) << name << " "
<< std::left << std::setw(8) << price << " "
<< std::setw(7) << quantity;
return (ss.str());
}
static std::string showHdr() {
std::stringstream ss;
ss << std::right << std::setw(14) << "Name" << " "
<< std::left << std::setw(8) << "Price" << " "
<< std::left << std::setw(7) << "Stock";
return (ss.str());
}
private:
std::string name;
double price;
int quantity;
};
Main now reduces to:
std::vector<Soda> machine;
// load the machine with soda's
machine.push_back(Soda("Cola", .80, 20));
machine.push_back(Soda("Root Beer", .75, 20));
machine.push_back(Soda("Lemon-Lime", .90, 20));
machine.push_back(Soda("Grape - Mango", .80, 20));
machine.push_back(Soda("Cream", .80, 20));
// show machine inventory
{
// hdr
std::cout << Soda::showHdr() << std::endl;
for (auto i : machine)
std::cout << i.show() << std::endl;
}