I am trying to understand the switch statement in C (I am using gcc from Ubuntu v16.04). I am able to understand its semantics, but have the following 2 questions about its syntax:
I have noticed after reading a few examples of usage of switch statement that the symbol after case is sometimes enclosed in '' and sometimes it isn't.
eg: case 1 or case 'a'.
I checked the Linux manpage for the switch statement (https://linux.die.net/man/1/switch) and there they have not used ' ' for a string. So I don't know what to do.
Sometimes the block of code within a single case is enclosed in a { } and sometimes it is not. I had read before that multi-line statements need to be enclosed in a { }, but not necessarily for single-line statements as in a for loop,while loop with single line statements etc. But sometimes a case statement has 1 line of code (for eg a *= 5;) followed by break statement (so total 2 statements) and yet both lines are not enclosed in { }. The Linux manpages haven't mentioned this. Can someone clarify this?
(1) 'a' is ascii value 97. Ascii is a standard way of encoding characters and it's used in many other languages as well. Essentially, each character is represented as a numerical value. So when you have:
...
case 'a':
...
you are actually executing the code below the case if the switch variable is equal to 97. In your example:
case '1':
checks if the switch variable is equal to char '1', which is ascii value 49.
(2) Enclosing a case statement with braces changes the scope of the variables between the braces. Consider the following example:
switch (sw) {
case 1:
int b = 2;
sw += b;
break;
case 2:
int b = 3;
sw += b;
break;
default:
break;
}
This is because in case 1 and case 2, you are instantiating an integer called "b". Since both case statements are in the same variable scope (the switch statement's scope), the compiler gives you an error since you are instantiating a variable with the same name and type twice. Now consider the code below:
switch (sw) {
case 1: {
int b = 2;
sw += b;
break;
} case 2: {
int b = 3;
sw += b;
break;
} default: {
break;
}
}
This code compiles. By enclosing each case's code in braces, you are giving each case its own variable scope, where it could redefine the same variable once in each scope.
To answer your questions:
q1: 1 and '1' are not the same. The latter is surrounded by single-quotes, which in C always represent a character. Depending on the C implementation, this character would be stored in the ASCII format, with the numerical representation 49. Seeing as a character in ASCII format is guaranteed the ability to be represented numerically, but a number isn't, the comparison '1' == 1 is legal, because the character will be implicity converted to an integer.
q2: Enclosing a case in curly braces is optional. You can declare a scope at any time using curly braces. More information: https://www.geeksforgeeks.org/scope-rules-in-c/, C Switch-case curly braces after every case
(1) You're asking about the switch statement in C (specifically, gcc), but the link you included is for the switch statement in a Linux shell. These are two different languages!
In C, the single quote '' is used for characters. A character is a single letter/number/symbol/etc, as opposed to a string which is one or more characters. So case 1: will match the number 1, and case '1': will match the character '1'. A number has a type like int or long. A character has the type char. So whether you use '' or not depends on if you're trying to match a character or a number.
(2) The { } are not necessary. You might choose to use them to visually group the block of code, but you don't have to. You also can use { } to limit the scope of a variable. Variable scope is a big topic, see this for more info: https://www.w3schools.in/c-tutorial/variable-scope/
There's no man page for the switch used in C. What you're looking at is the man page for the switch command (as per the (1) in the name), something totally different.
A case needs to use single quotes for a character constant and no single quotes for a non-character constant.
For example, case '1': checks for the character '1' (typically with an integer value of 49), and case 1: checks for the integer value 1.
The curly braces are mostly a preference of style. Some people think the curly braces makes their code look nicer and/or clearer, and some don't. There's no difference except for scope.
This allows you to define a variable within an individual case label so you don't make it available to the whole function or switch statement.
Related
I am tired of my university, my midterm exams and I am tired of thinking what the hell is wrong here.
double n;
n = 3.5; // line 2
switch( n){
case 2.5 : printf(“High”); break;
case 0.5 : printf(”Low”); break;}
In general, I thought default statement is missing because n is 3.5 and every switch statement we wrote included a default statement.
Options are:
I)default statement is missing (What I ticked.)
II)there is an error in line 2
III)semi colon is missing at the end of the switch statement
IV)double data type cannot be used with switch statement (Correct
answer.)
Can anyone explain me why IV is correct answer? Thank you.
The label default may be omitted in the switch statement. However the case labels shall be of integer type.
So the switch statement is incorrect at least due to incorrect case labels.
Also the expression in the switch statement shall have an integer or an enumeration type.
switch work on integral values (int, enum).
If you want to "switch" on double values, you will need to use if/else statements.
I'm new in C and I would like to know how can I correct this error I can keep getting please:
"expression must have integral type"
switch (detect_format(format))
{
case "%d":
printf("teqs");
break;
default:
break;
}
note : detect_format(format) returns a string.
As per the standard, C11, chapter 6.8.4.2:
The controlling expression of a switch statement shall have integer type
A string is not allowed.
That said, the case labels, also need to be integer constant expression, something like case "%d": is also illegal.
If you want to take a decision based on the returned string, you will need to use strcmp() and use the result as the controlling expression of a switch statement.
You Can't use switch-case with %d like you did in your code: (case "%d").
What you have to do is to use it with integers like:
case 1:
//code//
break;
case 2:
//code//
break;
//and so on...
there are smarter solutions but since you're new to C Programming, you can use simple integers to adjust your specific cases.
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'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.