int a = 10;
switch(a){
case 0:
printf("case 0");
break;
case 1:
printf("case 1");
break;
}
Is the above code valid?
If I am sure that int a will not have any other value than 1 and 0, can I avoid default?
What if in any case a value will be different from 1 and 0?
I know this is a silly question but I was thinking that perhaps it would be illegal or undefined behavior soI just asked to make sure.
The code is valid. If there is no default: label and none of the case labels match the "switched" value, then none of the controlled compound statement will be executed. Execution will continue from the end of the switch statement.
ISO/IEC 9899:1999, section 6.8.4.2:
[...] If no converted case constant expression matches and there is no default label, no part of the switch body is executed.
As others have pointed out it is perfectly valid code. However, from a coding style perspective I prefer adding an empty default statement with a comment to make clear that I didn't unintentionally forget about it.
int a=10;
switch(a)
{
case 0: printf("case 0");
break;
case 1: printf("case 1");
break;
default: // do nothing;
break;
}
The code generated with / without the default should be identical.
It is perfectly legal code. If a is neither 0 or 1, then the switch block will be entirely skipped.
It's valid not to have a default case.
However, even if you are sure that you will not have any value rather than 1 and 0, it's a good practice to have a default case, to catch any other value (although it is theoretically impossible, it may appear in some circumstances, like buffer overflow) and print an error.
Default is not mandatory, but it always good to have it.
The code is ideally, but our life is not, and there isn't any harm in putting in a protection there. It will also help you debugging if any unexpected thing happens.
Yes, the above code is valid.
If the switch condition doesn't match any condition of the case and a default is not present, the program execution goes ahead, exiting from the switch without doing anything.
It's same like no if condition is matched and else is not provided.
default is not an mandatory in switch case. If no cases are matched and default is not provided, just nothing will be executed.
The syntax for a switch statement in C programming language is as follows:
switch(expression) {
case constant-expression :
statement(s);
break; /* optional */
case constant-expression :
statement(s);
break; /* optional */
/* you can have any number of case statements */
default : /* Optional */
statement(s);
}
Related
I stumbled upon this code, which works as expected:
switch (ev->deviceType) {
break;
case DEVICE_TS1E0:
//some code
break;
case DEVICE_TS1E3:
//some code
break;
default:
//some logging
break;
}
Now, there's a lonesome break; at the start of the switch, which appears to have no effect.
Is there any circumstances where that break; would have an effect ?
TL;DR That break statement is ineffective and a dead-code. Control will never reach there.
C11 standard has a pretty good example of a similar case, let me quote that straight.
From Chapter §6.8.4.2/7, (emphasis mine)
EXAMPLE In the artificial program fragment
switch (expr)
{
int i = 4;
f(i);
case 0:
i = 17;
/* falls through into default code */
default:
printf("%d\n", i);
}
the object whose identifier is i exists with automatic storage duration (within the block) but is never
initialized, and thus if the controlling expression has a nonzero value, the call to the printf function will
access an indeterminate value. Similarly, the call to the function f cannot be reached.
That statement, along any other statements that is not in a case clause in switch statement are unreachable code, aka dead code. That means they will not be run anyway. It is not recommended to use them.
Into the following code case 1: to case 2: and case 5: seem to have no code for execution. My question is can't we just omit typing them?
switch(c)
{
case 1:
case 2:
case 3:
a++;
break;
case 5:
default:
b++;
break;
}
They fall through, i.e. if you encounter 1 or 2 it will jump into these cases and since there's no break just continue with case 3. So you can not omit those, because otherwise 1 and 2 would jump to default.
Similarily, case 5 will fall through to default, meaning that you could omit case 5.
Essentially, switch statements can be imagined as goto-jumps to the appropriate positions. That means, that the program will jump into the appropriate case and continue working from there until it finds break or return. This means, that you have to write break or return explicitly if you don't want your program to continue execution in the subsequent case.
No, you can't omit them. Basically cases 1 and 2 will fallthrough to 3. Omitting them would cause the default case to be executed for input values 1 and 2.
It's a fallthrough case. That means the cases fall down until a break appears. So a++ would be excecuted with c being 1, 2, 3.
The cases for 1,2 and 3 mean that if c=1,2 or 3 then a++. If c=5 or different then b++ - note that the code only stops if you hit a break so that c=1 gives the same as c=3 for example.
They may be omitted but the result would be different.
Basically what the code does is for c equal to 1, 2, or 3 a++ gets executed, otherwise, b++ does.
In this case case 5: should be omitted after all as it will falls in the default: case.
My question is can't we just omit typing them?
TL;DR -- In your case, No. They have some meaning.
Quoting C11, chapter §6.8.4.2, The switch statement
The integer promotions are performed on the controlling expression. The constant expression in each case label is converted to the promoted type of the controlling expression. If a converted value matches that of the promoted controlling expression, control jumps to the statement following the matched case label....
That means, based on the value of c here, the particular "case" will be executed.
Fine. Wait, now we know where to take the control to start execution, but where to stop, exactly?
You're thinking "before the next case body?"
Well, not exactly. It does not automatically stop . Usually, we use a break; statement after each case block, to "mark" the end of that particular case. In case, the break; is not present, it will continue to execute the statements from following cases (if any), as if they are part of the same case block and it will continue until it reaches the end of switch body.
Now, to elaborate, in your case, if the value of c is either 1, 2 or 3, it will execute the statement block for case 3. Notice, there is no break; statement after case 1: and case 2: labels. It is kind of "fall-through" technique.
OTOH, if you remove the case 1: and case 2: labels, if c holds 1 or 2, the control won't reach the block after case 3:, it will go to default: label.
However, case 5: will fall through to default: label, making it redundant. This once can be removed.
You can indeed omit a case. This switch statement
switch(c)
{
case 1:
case 2:
case 3:
a++;
break;
case 5:
default:
b++;
break;
}
can be rewritten like
switch(c)
{
case 1:
case 2:
case 3:
a++;
break;
default:
b++;
break;
}
without using label case 5: . But you may not exclude labels case 1: and case 2:
With these labels the switch statement can be rewritten using the if-else statement the following way
if ( 1 <= c && c <= 3 )
{
a++;
}
else
{
b++;
}
On the other hand without these case labels the corresponding if-else statement will look like
if ( c == 3 )
{
a++;
}
else
{
b++;
}
As you can see yourself there is a big difference between these two if-else statements.
Section 6.8.1 of C11 or C99, or section 3.6.1 of C89 all seem to indicate that default and case x (where x is some constant-expression) are examples of labeled statements, along-side identifier:-style labels that are suitable for use with goto.
I'm aware that I could simply place an identifier:-style label directly following the default: or case x: labels. That's not what this question is about. I'm more curious as to whether there is any actual rationale behind prohibiting this kind of behaviour.
If it were possible to declare default: labels outside of a switch selection structure, then I would understand, as there would be some conflict between where the goto inside of the switch selection structure is intended to aim. However, section 6.4.1 of C11 or C99 or 3.1.1 of C89 prohibits the use of default as anything other than a keyword, and 6.8.1 restricts its use further to switch structures only (or generic structures in C11, which are irrelevant here).
I would also understand if multiple (possibly nested) switch structures, each with default: (or case x:) labels introduced ambiguity, however the scope of those labels seems to be restricted to within their inner-most surrounding switch structures, and referring to any identifier outside of its scope is clearly an error requiring a diagnostic at compile-time.
Has this been discussed in any standard documents (e.g. the rationale)? Is there any kind of explanation for this behaviour other than "it is because it is" or "because the spec says so"? If so, what is that explanation?
(I don't see how goto to a case would work syntactically.)
As you say, case and default labels only have the scope of the corresponding switch statement and can only be jumped to from outside. On the other hand
labels in C have function scope and can be jumped to from anywhere in the function.
So we are talking of labels with a quite different properties, they are probably treated quite different, internally. It seems relatively complicated to reconcile these properties and would make implementing this more complicated. All of a sudden you would have to have to decide when implementing goto which case or default is the correct one, whereas now you just have to put the address of the file scope identifier, there.
All of this said, this is just a guess about the original intentions for this distinction. But I am sure that argumentation along these lines would quickly kill any attempt to introduce such a feature now.
counter example
fun() {
switch (b) {
case x:
doSomething();
break;
}
switch(b2) {
case x:
doMore();
break;
}
goto x; // which one?
}
ISO/IEC 9899:2011 Information technology -- Programming languages -- C
6.8.1 Labeled statements
Syntax
labeled-statement:
identifier : statement
case constant-expression : statement
default : statement
A case or default label shall appear only in a switch statement. Further constraints on such labels are discussed under the switch
statement.
Label names shall be unique within a function.
Any statement may be preceded by a prefix that declares an identifier as a label name. Labels in themselves do not alter the
flow of control, which continues unimpeded across them.
6.8.6 Jump statements
Syntax
jump-statement:
goto identifier ;
6.8.6.1 The goto statement
The identifier in a goto statement shall name a label located somewhere in the enclosing function.
A goto statement causes an unconditional jump to the statement prefixed by the named label in the enclosing function.
6.2.1 Scopes of identifiers
A label name is the only kind of identifier that has function scope. It can be used (in a goto statement) anywhere in the function in which it appears, and is declared implicitly by its syntactic appearance (followed by a : and a statement).
case, default labels are not identifiers, named labels.
"Because the spec says so." Do not overthink it.
I'm aware that I could simply place an identifier:-style label directly following the default: or case x: labels. That's not what this question is about. I'm more curious as to whether there is any actual rationale behind prohibiting this kind of behaviour.
At cost of sounding tautologic, the rationale is that each case is intended to be uniquely jumped into from one single goto (the one created by the switch statement). Which is also why we need to group together more case labels if we want a block of code to be executed in case of more than one condition.
The following code might give you a visual answer. If goto [case label] were allowed, what would you write in order to go to the first case, goto case 'q' or goto case 'e'? Which one is the boss of that block of code?
#include <stdio.h>
int yes_or_no (char input, char additional_yes) {
int retval;
switch (input) {
case 'q':
case 'e':
printf("Bye bye! :-)\n");
retval = 0;
break;
case 'n':
printf("I am very sad that you said no :-(\n");
retval = 0;
break;
case 'y':
other_yes:
printf("Thank you for saying yes! :-)\n");
retval = 0;
break;
case '\n':
case '\r':
case ' ':
case '\t':
case '\v':
case '\f':
printf("Mmmmm... maybe you need time to think...\n");
retval = 1;
break;
default:
if (input == additional_yes) {
goto other_yes;
}
printf("\"%c\" is not an answer!\n", input);
retval = 1;
}
return retval;
}
int main (int argc, char *argv[]) {
printf("Please, answer (yes/ja or no/nein) -- e or q to quit\n");
while (yes_or_no(getchar(), 'j'));
return 0;
}
Consider also that, while optimizing, the compiler might decide not to use jump tables at all if it thinks that a simple repeated if would be faster than a switch (this might happen especially when the case labels are not many). In this case, if goto [case label] were allowed, you would force the compiler to create a label where normally there wouldn't be one.
This is one of the things I like most of C: if you want something you have to say it.
I'm reading KN King's A Modern Approach to C Programming, 2nd edition.
It says, there are also other forms of switch statement besides general switch statement (with case keyword).
The general form of switch statement is
switch (exp)
{
case constant-exp:
statement;
break;
case constant-exp:
statement;
break;
...
...
default:
statement;
break;
}
It also says (in Q&A) switch statement can have form with no case keyword for example.
I tried running an example with no case keyword, but it doesn't run (under std=-c99).
So, I wanted to know what are the other forms of switch statement that are valid in Standard C99.
EDIT: Cited fro BOOK
In it's most common form, the switch statement has the form
switch ( expression ) {
case constant-expression : statements
...
case constant-expression : statements
default : statements
}
Q&A
**Q: The template given for the switch statement described it as the "most common form." Are there other forms?
A**: The switch statement is a bit more general than described in this chapter, although the description given here is general enough for virtually all programs.
For example, a switch statement can contain labels that aren't preceded by the word case, which leads to amusing (?) trap. Suppose that we accidentally missell the word default:
switch(...) {
...
defualt: ...
}
The compiler may not detect the error, since it assumes that defualt is an ordinary label.
The syntax of a switch statement is:
switch ( expression ) statement
where the statement portion is typically a block (compound statement) containing labeled statements of the form:
case constant-expression : statement
or
default : statement
A switch statement isn't required to contain case or default labels, but there's no point in using a switch if you're not going to have one or more such labels. For example, this:
switch (42);
is a perfectly legal switch statement (the controlled statement is the null statement ;), but it's also perfectly useless.
I suspect you've misunderstood what the book says.
Your quote from the book says:
For example, a switch statement can contain labels that aren't
preceded by the word case, which leads to amusing (?) trap. Suppose
that we accidentally mispsell the word default:
switch(...) { ... defualt: ... }
The contents of a switch statement should be a block containing a sequence of case and default labels, each one ending either with a break or with a comment indicating that the control flow falls through to the next case. The point is that the language doesn't require this; the way it specifies the syntax gives you a lot of freedom (perhaps too much!). The only restriction is that case and default labels cannot appear outside a switch statement.
For example, suppose you accidentally write:
enum blah { foo, bar, baz };
switch (expr) {
case foo:
/* ... */
break;
bar: /* forgot the `case` keyword */
/* ... */
break;
defualt: /* misspelled "default" */
/* ... */
break;
}
Neither bar: nor defualt: was what was intended -- but they're both perfectly legal. They're ordinary labels, the kind that can be the target of a goto statement. Since there is no goto targeting either label, the corresponding chunks of code will never be executed. If expr is equal to foo, it will jump to the case foo:; for any other value, it will jump to the end of the switch statement.
And because they're perfectly legal, a compiler won't necessarily warn you about the error.
This is a common phenomenon in C. The grammar is so "dense" that a seemingly minor typo can easily give you something that's syntactically valid, but whose behavior is entirely different from what you intended.
Crank up the warning levels on your compiler, and pay attention to all the warnings you see. And be careful; the responsibility for writing your code correctly is ultimately yours. The compiler can help, but it can't catch all errors.
Regarding your observation: there are also other forms of switch statement besides general switch statement (with case keyword) Generally, the switch statement is very well documented, but there are a few interesting variations in the way the case statements are used...
Although nothing Earth shaking here, It may be useful to note: Sun (a flavor of unix) and GNU C compiler have an extension that provides case ranges for use with the switch() statement. So, for example, rather than using the classic syntax:
:
case 'A':
case 'B':
:
case 'Z':
//do something here.
break;
and so on...
A case range syntax can be used to delineate the conditions:
switch(input) {
case 'A' ... 'Z':
printf("Upper case letter detected");
break;
case 'a' ... 'z':
printf("Lower case letter has been detected");
break;
};
Important Note:, case ranges are not part of the C standard (C99 or C11) rather only an extension of the environments I have mentioned, and in no way should be considered portable. Case ranges are gaining in popularity (or at least in interest) and may be included as part of the C standard at some point, but not yet (AFAIK).
The go-to source for C-99 is the C-99 standard, though of course C-99 has been replaced by C11. The switch statement is on page 134 of the C-99 standard. They give an example of what is probably the most non-general switch statement you can have:
EXAMPLE In the artificial program fragment
switch (expr)
{
int i = 4;
f(i);
case 0:
i=17;
/* falls through into default code */
default:
printf("%d\n", i);
}
the object whose identifier is i exists with automatic storage duration (within the block) but is never
initialized, and thus if the controlling expression has a nonzero value, the call to the printf function will
access an indeterminate value. Similarly, the call to the function f cannot be reached.
Note the ways that this is not "standard" (and is generally bad code).
You have code that is not under any case or default label. In fact, nothing but identifier declarations seems to be acknowledged, so that i is a valid variable within the scope of the switch statement, but setting it or calling functions without a case or default label causes errors.
What I think the author wanted you to notice was that it's valid to not have a break under each label. In this example, case 0 falls through to the default label, but if there were other case labels beneath case 0, it would go through each one until you did hit a break statement.
For that matter, though not in this example, you can put the default label first. Again, if you don't put a break after it, you'll execute code under any following labels.
As I mentioned in my comment, you could just have a default label if you want, but that effectively renders the switch statement meaningless:
switch (exp)
{
default:
statement;
}
that's effectively equivalent to { statement; }.
Incidentally, you can do some clever (but confusing) tricks with avoiding break statements, e.g. this is a valid (though less efficient than c - '0') way to convert a digit character c to an integer:
int i = 0;
switch (c) {
case '9': ++i;
case '8': ++i;
case '7': ++i;
case '6': ++i;
case '5': ++i;
case '4': ++i;
case '3': ++i;
case '2': ++i;
case '1': ++i;
case '0':
default:
}
Wiki describes how the switch statement defined as below considering different coding languages,
In most languages, a switch statement is defined across many individual lines using one or two keywords. A typical syntax is:
1. The first line contains the basic keyword, usually switch, case or select,
followed by an expression which is often referred to as the control expression
or control variable of the switch statement.
2. Subsequent lines define the actual cases (the values) with corresponding
sequences of statements that should be executed when a match occurs.
Each alternative begins with the particular value, or list of values, that the control variable may match and which will cause the control to go to the corresponding sequence of statements. The value (or list/range of values) is usually separated from the corresponding statement sequence by a colon or an implication arrow. In many languages, every case must also be preceded by a keyword such as case or when. An optional default case is typically also allowed, specified by a default or else keyword; this is executed when none of the other cases matches the control expression.
Normally when using a switch statement, you cannot define and initialize variables local to the compound statement, like
switch (a)
{
int b = 5; /* Initialization is skipped, no matter what a is */
case 1:
/* Do something */
break;
default:
/* Do something */
break;
}
However, since the switch statement is a statement like for or while, there is no rule against not using a compound statement, look here for examples. But this would mean, that a label may be used between the closing parenthesis after the switch keyword and the opening brace.
So in my opinion, it would be possible and allowed to use a switch statement like this:
switch (a)
default:
{
int b = 5; /* Is the initialization skipped when a != 1? */
/* Do something for the default case using 'b' */
break;
case 1: // if a == 1, then the initialization of b is skipped.
/* Do something */
break;
}
My question: Is the initialization necessarily performed in this case (a != 1)? From what I know of the standards, yes, it should be, but I cannot find it directly in any of the documents I have available. Can anyone provide a conclusive answer?
And before I get comments to that effect, yes, I know this is not a way to program in the real world. But, as always, I'm interested in the boundaries of the language specification. I'd never tolerate such a style in my programming team!
Most people think of a switch as a mutiple if, but it is technically a calculated goto. And the case <cte>: and default: are actually labels. So the rules of goto apply in these cases.
Your both your examples are syntactically legal, but in the second one, when a==1 the b initialization will be skipped and its value will be undefined. No problem as long as you don't use it.
REFERENCE:
According to C99 standard, 6.2.4.5, regarding automatic variables:
If an initialization is specified for the object, it is performed each time the declaration is reached in the execution of the block;
So the variable is initialized each time the execution flow reaches the initialization, just as it were an assignment. And if you jump over the initialization the first time, then the variable is left uninitialized.