FAQ: Conditionals & Logic - Switch Statement

This community-built FAQ covers the “Switch Statement” exercise from the lesson “Conditionals & Logic”.

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This exercise can be found in the following Codecademy content:

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FAQs on the exercise Switch Statement

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The content says: Note: Without the break keyword at the end of each case, the program would execute the code for all matching cases and the default code as well. This behavior is different from if / else conditional statements which execute only one block of code.

Is this correct? without the ‘break’ keyword, i don’t think the program executes the code for “all matching cases AND the default case”, rather, it simply executes “all the code within that matching case and below”. For example, if you let number = 1 in the example, it will execute cases of 1, 2, 3, 4, and 5 because case 5 has ‘break’ keyword in it. Note that number is clearly NOT 2 nor 3 nor 4 nor 5. so case 2, case 3, case 4, case 5 are NOT matching case at all.

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Hello @mraji. Welcome to the forum!
You are EXACTLY right! I believe this issue has already been reported to Codecademy. I’ll check for sure. If it hasn’t, I’ll report it.

Update: Didn’t see it in the que of active bugs, so I made a report. Thanks for pointing out this error.

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Can a switch statement be used together with relational operators such as: <, !=, >= please?
For example,

#include <iostream>

int main() {

    int rating = 4;

    switch (rating) {

        case (rating <= 3): // Is this possible please?
            std::cout << "Less than or equal to 3 stars!\n";
        break;
        default:
            std::cout << "Better than 3 stars!\n";
        break;

    }

}

Thank you!

Hello, @shenobey.
I don’t think you can use an expression like that in a case. Someone please correct me if I’m wrong. You can, however, use a range like this:

#include <iostream>

int main() {

    int rating = 4;

    switch (rating) {

        case 0 ... 3: // You have to include a space then 3 dots, so the compiler knows the dots aren't decimals. The space after the dots is optional. The values are inclusive.
            std::cout << "Less than or equal to 3 stars!\n";
        break;
        default:
            std::cout << "Better than 3 stars!\n";
        break;

    }

}

You can also use ranges of letters: 'A' ... 'M'. The spaces, I believe are optional with non-numeric values, so 'A'...'M' should also work.

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Thank you very much, midlindner!

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Why is this the case? In mraji’s example, why would the computer execute case 2 , 3, etc, if number = 1?

Like isn’t the computer supposed to only execute cases that match the condition?

int main() {
int number = ?;
switch(number) {
case 1 :
std::cout << “Bulbusaur\n”;
case 2 :
std::cout << “Ivysaur\n”;
case 3 :
std::cout << “Venusaur\n”;
break;
case 4 :
std::cout << “Charmander\n”;
break;
case 5 :
std::cout << “Charmeleon\n”;
break;
case 6 :
std::cout << “Charizard\n”;
break;
case 7 :
std::cout << “Squirtle\n”;
break;
case 8 :
std::cout << “Wartortle\n”;
break;
case 9 :
std::cout << “Blastoise\n”;
break;
default :
std::cout << “Unknown\n”;
break;
}
}

If i set number to 1 it will still run case 2 and 3 because they don’t have breaks, but if i set number to 3 it will only run case 3, presumably because the computer recognizes that number != case 1 or case 2. Why can the computer recognize the inequality in the latter case but not the former?

Hello @arnodunstatter.

Everything between the { }'s following the switch() statement is a single block of code. The case 1:, case 2:, etc. are just labels. Basically what happens is when the expression inside of the switch() statement’s parenthesis is evaluated, control flow is passed to the matching label. Labels aren’t executed. They just mark a spot in the switch statement’s code block. When control is passed to first matching case: label, every line of executable code that follows inside of the code block gets executed. If control reaches a break; statement it gets passed to the next line of code outside the switch statement’s code block.

1 Like