Reputation: 4305
In this loop (assuming amount
is 2), instead of printing:
1. [op1]
2. [op2]
it prints:
1. [op1]
2. [op2]
3. [op3]
4. [op4]
5. [op5]
Why? How can I fix this?
for(int i=1;i<=amount;i++){
//string s;
switch(i){
case 1:
cout << i << ". " << op1 << endl;
//s = op1;
case 2:
cout << i << ". " << op2 << endl;
//s = op2;
case 3:
cout << i << ". " << op3 << endl;
//s = op3;
case 4:
cout << i << ". " << op4 << endl;
//s = op4;
case 5:
cout << i << ". " << op5 << endl;
//s = op5;
}
//cout << i << ". " << s << endl;
}
Upvotes: 1
Views: 618
Reputation: 70186
You need to use break;
or switch()
will have what's called "fallthrough"
.
(This means that it executes your case: and then continues on anything that comes afterwards, including code in other case statements).
Upvotes: 10
Reputation: 1344
Put a "break" statement after the cases, otherwise the control flow will just go straight through.
Upvotes: 1
Reputation: 32258
You have to insert break;
-statements at the end of each of your case:
-parts.
This is because switch () {}
is merely a reformulation of a goto
, the cases are just labels, not real blocks (that's why they use the label-syntax instead of a block-syntax like case ... { }
.
Upvotes: 2
Reputation: 389
You need to put
break;
at the end of each case. Otherwise it goes to case1, and then continues through to the end of the select. (case1, 2, 3, 4, 5)
Upvotes: 2