问题
Through a little typo, I accidentally found this construct:
int main(void) {
char foo = 'c';
switch(foo)
{
printf("Cant Touch This\n"); // This line is Unreachable
case 'a': printf("A\n"); break;
case 'b': printf("B\n"); break;
case 'c': printf("C\n"); break;
case 'd': printf("D\n"); break;
}
return 0;
}
It seems that the printf at the top of the switch statement is valid, but also completely unreachable.
I got a clean compile, without even a warning about unreachable code, but this seems pointless.
Should a compiler flag this as unreachable code?
Does this serve any purpose at all?
回答1:
Perhaps not the most useful, but not completely worthless. You may use it to declare a local variable available within switch scope.
switch (foo)
{
int i;
case 0:
i = 0;
//....
case 1:
i = 1;
//....
}
The standard (N1579 6.8.4.2/7) has the following sample:
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
iexists with automatic storage duration (within the block) but is never initialized, and thus if the controlling expression has a nonzero value, the call to theprintffunction will access an indeterminate value. Similarly, the call to the functionfcannot be reached.
P.S. BTW, the sample is not valid C++ code. In that case (N4140 6.7/3, emphasis mine):
A program that jumps90 from a point where a variable with automatic storage duration is not in scope to a point where it is in scope is ill-formed unless the variable has scalar type, class type with a trivial default constructor and a trivial destructor, a cv-qualified version of one of these types, or an array of one of the preceding types and is declared without an initializer (8.5).
90) The transfer from the condition of a
switchstatement to a case label is considered a jump in this respect.
So replacing int i = 4; with int i; makes it a valid C++.
回答2:
Does this serve any purpose at all?
Yes. If instead of a statement, you put a declaration before the first label, this can make perfect sense:
switch (a) {
int i;
case 0:
i = f(); g(); h(i);
break;
case 1:
i = g(); f(); h(i);
break;
}
The rules for declarations and statements are shared for blocks in general, so it's the same rule that allows that that also allows statements there.
Worth mentioning as well is also that if the first statement is a loop construct, case labels may appear in the loop body:
switch (i) {
for (;;) {
f();
case 1:
g();
case 2:
if (h()) break;
}
}
Please don't write code like this if there is a more readable way of writing it, but it's perfectly valid, and the f() call is reachable.
回答3:
There is a famous use of this called Duff's Device.
int n = (count+3)/4;
switch (count % 4) {
do {
case 0: *to = *from++;
case 3: *to = *from++;
case 2: *to = *from++;
case 1: *to = *from++;
} while (--n > 0);
}
Here we copy a buffer pointed to by from to a buffer pointed to by to. We copy count instances of data.
The do{}while() statement starts before the first case label, and the case labels are embedded within the do{}while().
This reduces the number of conditional branches at the end of the do{}while() loop encountered by roughly a factor of 4 (in this example; the constant can be tweaked to whatever value you want).
Now, optimizers can sometimes do this for you (especially if they are optimizing streaming/vectorized instructions), but without profile guided optimization they cannot know if you expect the loop to be large or not.
In general, variable declarations can occur there and be used in every case, but be out of scope after the switch ends. (note any initialization will be skipped)
In addition, control flow that isn't switch-specific can get you into that section of the switch block, as illustrated above, or with a goto.
回答4:
Assuming you are using gcc on Linux, it would have given you a warning if you're using 4.4 or earlier version.
The -Wunreachable-code option was removed in gcc 4.4 onward.
回答5:
Not only for variable declaration but advanced jumping as well. You can utilize it well if and only if you're not prone to spaghetti code.
int main()
{
int i = 1;
switch(i)
{
nocase:
printf("no case\n");
case 0: printf("0\n"); break;
case 1: printf("1\n"); goto nocase;
}
return 0;
}
Prints
1
no case
0 /* Notice how "0" prints even though i = 1 */
It should be noted that switch-case is one of the fastest control flow clauses. So it must be very flexible to the programmer, which sometimes involves cases like this.
回答6:
It should be noted, that there are virtually no structural restrictions on the code within the switch statement, or on where the case *: labels are placed within this code*. This makes programming tricks like duff's device possible, one possible implementation of which looks like this:
int n = ...;
int iterations = n/8;
switch(n%8) {
while(iterations--) {
sum += *ptr++;
case 7: sum += *ptr++;
case 6: sum += *ptr++;
case 5: sum += *ptr++;
case 4: sum += *ptr++;
case 3: sum += *ptr++;
case 2: sum += *ptr++;
case 1: sum += *ptr++;
case 0: ;
}
}
You see, the code between the switch(n%8) { and the case 7: label is definitely reachable...
* As supercat thankfully pointed out in a comment: Since C99, neither a goto nor a label (be it a case *: label or not) may appear within the scope of a declaration that contains a VLA declaration. So it's not correct to say that there are no structural restrictions on the placement of the case *: labels. However, duff's device predates the C99 standard, and it does not depend on VLA's anyway. Nevertheless, I felt compelled to insert a "virtually" into my first sentence due to this.
回答7:
You got your answer related to the required gcc option -Wswitch-unreachable to generate the warning, this answer is to elaborate on the usability / worthyness part.
Quoting straight out of C11, chapter §6.8.4.2, (emphasis mine)
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
iexists with automatic storage duration (within the block) but is never initialized, and thus if the controlling expression has a nonzero value, the call to theprintffunction will access an indeterminate value. Similarly, the call to the functionfcannot be reached.
Which is very self-explanatory. You can use this to define a locally scoped variable which is available only within the switch statement scope.
回答8:
It is possible to implement a "loop and a half" with it, although it might not be the best way to do it:
char password[100];
switch(0) do
{
printf("Invalid password, try again.\n");
default:
read_password(password, sizeof(password));
} while (!is_valid_password(password));
来源：https://stackoverflow.com/questions/41727415/valid-but-worthless-syntax-in-switch-case