Switch in C++ skips all cases and goes straight t

Question

I'm new to this site but have been taking a programming class for about a semester now. I just had a question as to why my program is skipping all the cases, goes to default, and closes immediately without stopping or doing anything. It's probably just a small thing but I've been sitting here forever and can't see it. Here's the code. Thanks for the help:

//Name Game
#include <iostream>  
#include <fstream>  
#include <stdlib.h> 
#include <math.h>  
#include <iomanip>  
using namespace std; 
int main(void)
{
    char name, zzz;
    cout << "Hello, and welcome to the start of a fantastic new adventure. \nI'm your guide for the day: Sebastian.  \nMay I ask what your name is?\n";
    cin >> name;
    switch (name)
    {
    case 'alex':
    case 'Alex':
        cout << "What an absolutely beautiful name. It sends chills down my spine just thinking about it. \nYou're one lucky girl.";
        cout << "\nI've heard from my friend that you are as beautiful as your name suggests.";
        break;
    case 'Ryan':
    case 'ryan':
        cout << "Please stop using this project.  What are you even doing here?";
        break;
    default:
        cout << "That's a nice name.  You should keep it for as long as you can.";
        cin >> zzz;
        return (0);
    }
    return 0;
}

You can not do that: Read the compiler warnings (errors) and a book. A case label must be an integral (integer) constant. — , Apr 21 '16 at 18:10
@DieterLücking, `'alex'` _is_ an integral constant (it's a _multicharacter literal_, which is conditionally-supported, has type `int`, and has an implementation-defined value) — Jonathan Wakely, Apr 21 '16 at 18:11

score 6 · Answer 1 · answered Apr 21 '16 at 18:09

6

A char is only a single character, not a whole name.

When you read a char, it will be a or A or ,, but never Alex.

You have to use strings for this task, which are written in "double" quotes, not 'single' quotes.

answered Apr 21 '16 at 18:09

Roland Illig

40,703
10
88
121

8

And you cannot switch on strings – Ed Heal Apr 21 '16 at 18:09

score 0 · Answer 2 · answered Apr 21 '16 at 18:19

0

I'm a bit surprised this compiles as the cases in the switch statement use single quotes to enclose multiple characters. Single quotes typically contain just a single character, to represent multiple characters (a 'string') you would use double quotes.

The name variable is declared as a char which can only hold a single character - I think this is why it's always selecting the default case.

I'd recommend reading up on handling strings in c++ to accomplish what you're trying to do.

answered Apr 21 '16 at 18:19

grae22

104
11

Ok, thanks you guys. I'll definitely make sure to read up on the difference between the two (my teacher never taught us strings so I'm on my own there). Anyway, thanks a lot! Appreciate ya'll! – Russell Lochrie Apr 25 '16 at 14:09

score 0 · Answer 3 · edited Nov 10 '16 at 04:15

Your problem is because you misunderstood variable types between char and string, where string is bunch (group) of char.

String name1 = "Alex";
cout<<name1;

it will print Alex (notice the double quotes), while

char name2 = 'Alex';
cout<<name2;

will print A, since variable name2 is only for one character

Because your case condition is multiple char and your switch variable is only one character, the switch-case statement goes to default.

The alternative is using string variable and change the switch-case statement to if-else statement since switch case cant handle string variable (without creating pointer or hash)

Switch in C++ skips all cases and goes straight t

3 Answers3