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.
Related
I recently came across the code snippet shown below, I was expecting it to be a syntax error but to my surprise, the code produces valid output.
#include <stdio.h>
int main(void) {
int x = 2;
switch(x) {
case 1: printf("1"); break;
do {
case 2: printf("2 "); break;
case 3: printf("3 "); break;
} while(++x < 4);
case 4: printf("4"); break;
}
return 0;
}
output: 2 4
Compiler: GCC 6.3
I found a similar problem but it is not justifying above condition completely,
Mixed 'switch' and 'while' in C
Can anyone explain,
What exactly happening here?
Why isn't it a syntax error?
Why case '3' is skipped?
case X: some_statement; is a labeled statement (6.8.1) just like goto_label: some_statement; with the only caveat that case/default labels may only appear inside the body of a switch (possibly in an arbitrarily nested compound statement). That makes case statements only very loosely coupled with switches, syntactically.
Semantically, switches are implementable as computed gotos and like regular gotos, they may jump pretty much anywhere (in C11, you can't jump past a VLA declaration) including inside of a loop (see https://en.wikipedia.org/wiki/Duff%27s_device#Mechanism for another description).
In your example, case 3: is skipped because of the break, but case 4: does follow because the break after case 3: is a loop-breaking break, not a switch-breaking break (break/continue always apply to the nearest construct they can apply to).
The official C grammar for a switch statement is:
switch ( expression ) statement
Any statement can be a labeled-statement, for which the grammar includes:
case constant-expression : statement
This means you can put pretty much whatever you want inside the body of the switch: It can be a compound statement which includes multiple statements and a do-while statement that includes multiple statements, and case labels can be prefixed to any of those statements.
The compiler merely implements the switch using jump instructions (inside its abstract machine; they may end up as other instructions after optimization). Loops statements such as do-while are implement with code that tests the controlling expression and jumps conditionally. So, in spite of the nice structure you think of in structured languages, it boils down to jump instructions, and those can be interwoven as desired.
The jumps are not the most disconcerting part of this. Object initialization and lifetime is more of a concern. Switch statements can unintentionally skip the initialization of objects if care is not taken.
The printf of “3 ” is never executed because control jumps from the switch to case 2, then breaks, which exits the do-while. This leaves code at the case 4 statement, which prints “4 ” and then breaks out of the switch statement.
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.
I'm using Keil C on an 8051 project and I'm getting error C175 - Duplicate Case value from the following:
switch (x)
{
case 0:
break;
case 1:
break;
case 2:
switch (y)
{
case 0:
break;
case 1:
break;
}
}
Any thoughts ?
Your compiler is broken. Keil is infamous for its poor standard compliance.
C11 6.8.4.2/3
The expression of each case label shall be an integer constant
expression and no two of the case constant expressions in the same
switch statement shall have the same value after conversion. There may
be at most one default label in a switch statement. (Any enclosed
switch statement may have a default label or case constant expressions
with values that duplicate case constant expressions in the enclosing
switch statement.)
The above bold, normative text is there in any version of the C standard.
This is likely a compiler bug. With gcc your code works flawlessly: http://ideone.com/u4svzf. This is the tested code:
#include <stdio.h>
int main(void) {
int x, y;
switch (x) {
case 0:
break;
case 1:
break;
case 2:
switch (y) {
case 0:
break;
case 1:
break;
}
}
// your code goes here
return 0;
}
You cannot put "0" and "1" as case statements within another switch which has 0 and 1 for a case.
You can find the explanation here.
[EDİT]: Actually the explanation says "in the same statement" but we understand from here: in keil C we cannot enclose a switch block in a switch block. Like #niklasfi said, it runs perfectly in gcc(I tried too).
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.
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);
}