I am looking through someone's C code, and discovered something I didn't even know was possible. Inside some of the cases, the switch variable is modified, with the intention that another case is executed after the break.
The switch is inside an interrupt handler, so it will get called repeatedly.
switch (variable)
{
case 1:
some_code();
variable = 3;
break;
case 2:
more_code();
variable = 5;
break;
case 3:
more_code();
variable = 5;
break;
case 4:
my_code();
break;
case 5:
final_code();
break;
}
The code seems to be working as the author intended.
Is this guaranteed behaviour in C? I always assumed that the break statement would cause execution to jump to directly after the switch statement. I wouldn't have expected it to continue testing every other case.
This is a common technique for state machine type code. But it doesn't jump automatically like you imagine. The switch statement has to be put inside a while loop. In this case, I imagine the loop would look something like:
while (!done) {
done = (variable == 5);
switch (variable) {
/*...*/
}
}
Alternatively, the switch statement could be inside a function body, and that function is called from a loop until the done condition is met.
Related
I was musing over a switch statement I had written for 4 mutually exclusive cases. I inserted a break statement at the end of each case, because I didn't want to do the test again after one case or another had been successful.
Here's the question, though. What, if anything, does the last break do? If the test case 43 succeeds, the break terminates the case, but if there's no break, the default shouldn't run, and so there's no wasted test anyway.
Here's the code:
switch(telemetry) {
case(40):
printf("\nHouse Telemetry #%i \n", psc);
break;
case(41):
printf("\nNav Telemetry #%i \n", psc);
break;
case(42):
printf("\nDownhill Telemetry #%i \n", psc);
break;
case(43):
printf("\nRealTime Telemetry #%i \n", psc);
break; // what do I do?
default:
printf("\nCommand ID not recognized\n");
}
Perhaps one of you compiler gurus can advise me.
If you did not have the break; in case 43, it would fall through to the default case, and also execute the printf("\nCommand ID not recognized\n");.
Why do you think the default would not run? switch-case will continue execution until it encounters a break statement, and this includes the default section.
The break therefore is required for the desired behavior.
Putting a break; on the very last element of a switch-case (whether it is default or case) is optional, as it would stop execution that is coming to an end anyway.
I've encountered programmers who feel both ways about it: That there should not be an extra break at the very end of the block because it is not necessary, or that there should be one for consistency, and on the possibility that the sections get re-arranged in the future.
A switch statement is not a language construction that divides codes into separate cases. It is a computed “go to” statement: Based on the switch expression, it causes program control to jump to one of the labels.
Once that jump is completed, the switch does not exercise any further control over execution. It does not separate cases from one another or insert jumps or returns at the end of each case. Therefore, if you want control to leave the code for one case and not flow into the following code for another case, you must insert a break or other instruction that affects program control (such as return).
This characteristic of switch statements makes possible uses such as:
switch (letter)
{
case 'a':
case 'e':
case 'i':
case 'o':
case 'u':
printf("Vowel.\n");
break;
default:
printf("Consonant.\n");
break;
}
Observe there are five case labels using one section of code. If the switch automatically separated them, the first four would be empty; nothing would be done for them when letter was a, e, i, or o.
Another example is where one case does some preparation work and then falls into another case, such as:
switch (operation)
{
case Subtract:
b = -b;
case Add:
result = a+b;
break;
case Multiply:
result = a*b;
break;
}
Here the Add case adds two numbers, and the Subtract case works by negating b and then continuing into the code for the Add case. This could not work if the switch automatically separated cases. (This is a simplified example, of course.)
The break keyword in each case indicates the end of a particular case. If we do not put the break in each case then even though the specific case is executed, the switch in C will continue to execute all the cases until the end is reached. This should not happen; hence we always have to put break keyword in each case. Break will terminate the case once it is executed and the control will fall out of the switch.
I'm working on an embedded system where a register hast to be accessed and after that to be incremented to achieve the result I'm looking for, since the machine is notifying and configured to react on my access and changing or not changing the flag. So the switch's argument has to be kept as it is, since it would otherwise change the behaving of the embedded system.
But there may occur a situation where I don't want to get any of the cases get invoked. But I still need to acces and increment the argument of the switch.
(More indepth I'm converting a sequence of analog values to digital values conversions step by step. The index is used to stay synchronized with the current conversion and relating it with the corresponding case to handle the figure correct. There may occur a state in which the index desynchronisizes to the current conversion so the sequence of conversions must be run through without any of the cases getting invoked (to prevent setting wrong data) untill the sequence is finished and the resynchroinisation can get performed)
The way I'm currently doing this is this:
switch (RunIndex++)/*RunIndex may only be accessed one time per execution
of this construct and has to be incremented in the same step. thats given.*/
{
if (RunIndexSyncedWithADCensured == false)
{
break;
}
case 0:
Case0RelatedData = SomeOperationsForCase0(RawData);
break;
case 1:
Case1RelatedData = SomeOperationsForCase1(RawData);
break;
case 2:
Case2RelatedData = SomeOperationsForCase2(RawData);
break;
default:
RunIndex = 0;
break;
}
This construct does the job but it looks like it is a bit controversial and I don't feel well by considering about committing this into productinal code.
So is there a better looking way to achieve the same, without the need of additional variables or assignements?
note:
Also it may be relevant, that this is in the first part of a interupt function consisting of 2 parts.
The first part handles what has to happen if() a conversion is finished. The second part, what has additional to be done if() this conversion also ended the sequence. So it is no option to simply return from the function without getting into the second part. and there is currently no loop structure where an if(...)break; may break out. (What is also the reason why I'm putting the if inside the switch scope, as it is at least by standard a valid way to break out.)
Firstly, the if() inside switch() will never be executed.
Consider the below code snippet:
#include <stdio.h>
int main(int argc, char *argv[])
{
int i = 2;
switch(i) {
if (i == 2) {
printf("I M HERE\n");
}
case 1:
printf("1\n");
break;
case 2:
printf("2\n");
break;
default:
printf("default\n");
break;
}
return 0;
}
For your code: you expect the string I M HERE to be printed. But that is not the case.
The output for the above code snippet is:
2
No statements before case/default(switch constructs): is executed inside switch
Now to answer for
I don't want to get any of the cases get invoked. But I still need to acces and increment the argument of the switch
Just move the if() outside to the switch().
if (RunIndexSyncedWithADCensured) {
switch (RunIndex++) {
case 0:
Case0RelatedData = SomeOperationsForCase0(RawData);
break;
/* Other cases here */
default:
RunIndex = 0;
break;
}
} else
RunIndex++;
Why not save the value first and then increment it and use the saved value in the switch? By the way this also includes two accesses, first to read the value from RunIndex and the second to increment it.
int runIndex = (RunIndex++);
if (RunIndexSyncedWithADCensured )
{
switch (runIndex)/*RunIndex may only be accessed one time per execution
of this construct and has to be incremented in the same step. thats given.*/
{
case 0:
Case0RelatedData = SomeOperationsForCase0(RawData);
break;
case 1:
Case1RelatedData = SomeOperationsForCase1(RawData);
break;
case 2:
Case2RelatedData = SomeOperationsForCase2(RawData);
break;
default:
RunIndex = 0;
break;
}
}
Since you are using adjacent index numbers, you could make an array of function pointers to replace the switch. That's what the optimizer will turn the switch into anyhow. So instead of your obscure switch, you get this:
if (RunIndexSyncedWithADCensured)
{
SomeOperationsForCase[RunIndex](RawData);
}
RunIndex++;
if (RunIndex > MAX)
{
RunIndex = 0;
}
Completely unrelated to the switch statement design: in case RunIndex is a sensitive volatile variable such as some hardware register, then you shouldn't use it directly in any form of computations. Make a copy of it:
volatile int RunIndex;
...
int index = RunIndex; // read from RunIndex
if (RunIndexSyncedWithADCensured)
{
SomeOperationsForCase[index](RawData);
}
index++;
if (index > MAX)
{
index = 0;
}
RunIndex = index; // write to RunIndex
This is standard practice for all such volatile variables.
I have below switch case in my code.
switch(condition)
case 'A' :
//Code part A
break;
case 'B' :
//Code part A
//Code part B
break;
case 'C' : //Some code
break;
code Part A is repeated in both case 'A' and case 'B'. I want to avoid duplication of code.
If we use fall though then we need to add an if condition for case B. Is this the only way for avoiding repetition of code?
If the order is not important, you can simply do:
switch (condition)
{
case 'B':
// Code part B
// no break
case 'A':
// Code part A
break;
...
}
A case 'B' will continue to execute through the case 'A' code because you didn't call a break.
Manipulating a switch statement to reduce duplication of code may work at first, but then you may add additional cases to the switch later, which may break that cleanness of that optimization. For example:
switch(condition)
case 'A' :
// Code part A
break;
case 'B' :
// Code part A
// Code part B
break;
case 'C' :
// Code part C
break;
case 'D' :
// Code part A
// Code part D
break;
Suddenly an optimization which seemed nice at the time, starts to become difficult to maintain, difficult to read and error prone.
Having already determined that there is common code, the cleanest response in my view is to write functions to perform the common code and call from each case. Going forward, this will continue to be maintainable.
Unfortunately, that's the only way, short of defining a function for partA.
You can reduce nesting by exiting the switch from inside the combined case label to make the code look a little more uniform:
switch (someValue) {
case 'A':
case 'B':
// Code part A
if (someValue == 'A') break;
// Code part B
break;
case 'C':
break;
}
This lets your part A and part B code have the same level of nesting.
Can "//Code part B" be executed before "//Code part A"? If so, you could just reorder them and let it fall through without an if condition.
I don't think there's much else to do, otherwise. One of the reasons for the creation of object-oriented languages was avoiding the duplication of code you have in imperative languages.
I was trying this :
#include<stdio.h>
int main() {
int i = 2;
switch(i) {
default:{
printf("Hi\n");}
case 1:
printf("Hi1\n");
case 2:
printf("Hi2\n");
}
}
output is "Hi2" as expected, however if i = 3,
#include<stdio.h>
int main() {
int i = 3;
switch(i) {
default:{
printf("Hi\n");}
case 1:
printf("Hi1\n");
case 2:
printf("Hi2\n");
}
}
Output is
"Hi"
"Hi1"
"Hi2"
How does program enter other cases which do not match? I know putting break in default would solve this.
Why this behavior? Is there anything mentioned in C specification for this?
In C (and many other languages) the cases are simply labels that get 'jumped' to. Once execution starts in a selected case, it flows just like normal. If you want execution to 'stop' at the end of a case 'block' you have to use the break statement (or some other flow control statement):
switch(i) {
default:{
printf("Hi\n");}
break;
case 1:
printf("Hi1\n");
break;
case 2:
printf("Hi2\n");
break;
}
}
For whatever it's worth, in my opinion this is was an unfortunate decision made by the language designers since falling through to the next case after execution one or more statements in a case are executed is very, very rarely desired. However, that's the way the language works.
C# addresses this by making it so falling out of a case is illegal - some sort of explicit flow control (a break or a goto) is required at the end of a sequence of of statements in a case (unless it's the last case in the switch).
This is because the code steps through each instruction unless explicitly stated not to.
In a switch() { }, you must be explicit.
Think about the instructions backed by this C. It would be a jump table. Without break, there would be no jump beneath each branch to go after the switch case.
Complementing #alex, try this.
#include<stdio.h>
int main() {
int i = 2;
switch(i) {
default:{
printf("Hi\n");}
case 2:
printf("Hi2\n");
case 1:
printf("Hi1\n");
}
}
I want to do something like:
case someenumvalue || someotherenumvalue:
// do some stuff
break;
Is this legal in C?
The variable that I am doing a switch on is an enumerated list data struct.
You can rely on the fact that case statements will fall-through without a break:
case SOME_ENUM_VALUE: // fall-through
case SOME_OTHER_ENUM_VALUE:
doSomeStuff();
break;
You can also use this in a more complicated case, where both values require shared work, but one (or more) of them additionally requires specific work:
case SOME_ENUM_VALUE:
doSpecificStuff();
// fall-through to shared code
case SOME_OTHER_ENUM_VALUE:
doStuffForBothValues();
break;
Yes, you can simply skip the break in the first case to achieve the fall-through effect:
switch (expr) {
case someenumvalue: // no "break" here means fall-through semantic
case someotherenumvalue:
// do some stuff
break;
default:
// do some other stuff
break;
}
Many programmers get into the trap of fall-through inadvertently by forgetting to insert the break. This caused me some headaches in the past, especially in situations when I did not have access to a debugger.
you need:
case someenumvalue:
case someotherenumvalue :
do some stuff
break;
You can use fallthrough to get that effect:
case someenumvalue:
case someotherenumvalue :
do some stuff
break;
A case statement like a goto -- your program will execute everything after it (including other case statements) until you get to the end of the switch block or a break.
As others have specificied, yes you can logically OR things together by using a fall through:
case someenumvalue: //case somenumvalue
case someotherenumvalue : //or case someothernumvalue
do some stuff
break;
But to directly answer your question, yes you can do a logical, or bit-wise or on them as well (it's just a case for the result of the operation), just be careful that you're getting what you'd expect:
enum
{
somenumvalue1 = 0,
somenumvalue2 = 1,
somenumvalue3 = 2
};
int main()
{
int val = somenumvalue2; //this is 1
switch(val) {
case somenumvalue1: //the case for 0
printf("1 %d\n", mval);
break;
case somenumvalue2 || somenumvalue3: //the case for (1||2) == (1), NOT the
printf("2 %d\n", mval); //case for "1" or "2"
break;
case somenumvalue3: //the case for 2
printf("3 %d\n", mval);
break;
}
return 0;
}
If you choose to do the second implementation keep in mind that since you're ||'ing things, you'll either get a 1 or 0, and that's it, as the case.