CODEWITHSUNDEEP
c++
attain
attain

Reputation: 3

Can I select a case label in a switch statement and then shows it on the screen in C++?

everyone! I think it could be a silly question, but I was wondering if it is possible to show a case label in a switch statement in C++. This is the statement:

switch(TYPE)
{
  case 'A': 
    (cost <= 30000);
    break;
    
  case 'B':
    (30000 < cost <= 60000);  
    break;
    
  case 'C':
    (cost > 60000);  
    break;  
}

In this code, the user writes the cost at the beginning of it. The ranges that are showed in each case are the prices of a product. Let's say they write 62000, so given that it costs more than 60000 it should go to the type C product. So I would like to show the type to the user, and I write:

cout <<"TYPE: "<< TYPE <<endl;

But when I run the code, this line appears empty. I would like to find the way to put the 'C' there, which is the type that corresponds to the cost. Or at least I would like to know if this is not possible so I can think of another way to make this happen. I appreaciate your time and consideration. :)

Upvotes: 0

Views: 118

Answers (2)

0x5453
0x5453

Reputation: 13589

It sounds like your logic might be a bit backwards. switch is used to check against the value of a variable that has already been set. Additionally, it can only check a single value per case; it cannot check a range.

If I interpret your question correctly, I think you want something like this:

if (cost <= 30000) {
    TYPE = 'A';
} else if (cost <= 60000) {
    TYPE = 'B';
} else {
    TYPE = 'C';
}

Upvotes: 2

Remy Lebeau
Remy Lebeau

Reputation: 597205

You are going about this backwards. A switch will not accomplish what you want. You need to use if..else instead, eg:

cout << "TYPE: ";

if (cost <= 30000)
    cout << 'A': 
else if (cost <= 60000)
    cout << 'B': 
else
    cout << 'C':

cout << endl;

Which you can take a step further by wrapping the logic inside a function, eg:

char getTYPE(int cost)
{
    if (cost <= 30000)
        return 'A': 
    else if (cost <= 60000)
        return 'B': 
    else
        return 'C':
}

cout << "TYPE: " << getTYPE(cost) << endl;

Upvotes: 5

Related Questions