If ... else switch ... statement

Question

Following code compiles on my gcc 5.4.0, doesn't generate any warnings and works fine:

if (a == 0) {
    puts("0");
} else switch (a) {
    case 1: puts("1"); break;
    case 2: puts("2"); break;
    default: puts("default"); break;
}

Is if ... else switch ... correct statement?

Similar question in C++: https://stackoverflow.com/questions/35985592/is-if-condition-try-legal-in-c/35985655#35985655 — Mohit Jain, Jul 05 '17 at 13:39
Related fun fact: The C language actually does not have an `else if` statement. It is an `else` statement without braces followed by another `if` statement. `else if(a==b){ }` is a brace-less, strangely indented version of `else /* new line */ { if(a==b){} }`. — Lundin, Jul 05 '17 at 15:13
@Lundin That's right. IMO many books and tutorials suggest that `else if` is a statement but actually it is not — Piotr, Jul 05 '17 at 15:27

score 9 · Accepted Answer · answered Jul 05 '17 at 13:32

There's nothing wrong with your code. The grammar for if ... else is as follows:

attr(optional) if ( condition ) statement-true else statement-false

switch is a statement, so it is allowed to go after else.

Note that else if is not a special construct either, it's just an if statement after an else.

Attie · Answer 2 · 2017-07-05T13:59:06.800

What you have is syntactically correct... but it does not make easy reading.

You will likely be pulled up on it if your code is being reviewed - because you're making use of an "un-braced, multi-line statement as part of a conditional".

Prefer to be explicit and write it like so:

if (a == 0) {
    puts("0");
} else {
    switch (a) {
        case 1: puts("1"); break;
        case 2: puts("2"); break;
        default: puts("default"); break;
    }
}

It is legal in the same way that the following is legal:

if (a == 0)
    puts("0");
else
    puts("not0");

Such constructs can lead to mistakes when re-visiting the code... I seem to remember that one of the recent "popular" vulnerabilities was implemented (by mistake... hopefully) in part due to this "un-braced" use of if.

score 3 · Answer 3 · answered Jul 05 '17 at 13:32

3

The switch is a statement, so this is just putting a statement in the else.

It's no different from

else
  a = 0;

or

else
{
  switch(a)
  {
  case 1:
  ...
  }
}

It's not a very common way of writing it, but it's fine.

answered Jul 05 '17 at 13:32

unwind

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score 3 · Answer 4 · answered Jul 05 '17 at 13:34

Nothing wrong with your code, that is the same of writing this:

if (a == 0) {
    puts("0");
} else {
    switch (a) {
        case 1: puts("1"); break;
        case 2: puts("2"); break;
        default: puts("default"); break;
    }
}

That is perfectly legit. else switch is not a real statement.

Your statement is

if(condition){code block} else {code block}

code block can contain any statement, such as your switch statement.

score 2 · Answer 5 · answered Jul 05 '17 at 13:33

Syntax of if is:

if ( condition ) statement-true  else statement-false
condition is an expression convertible to a boolean (true/false)
statement-true is a statement which is executed if condition is true
statement-false is a statement which is executed if condition is false

And switch block is a statement.

So yes, your code exhibit correct statement.

If ... else switch ... statement

5 Answers5