How to handle when there is a new SW Release sometime and it adds another index to switch case. Index represents a parameter in this case. For example,
Rel1: i = 1-5, 7 (excluding 6)
Rel2: i = 1-7
for (int i = 1; i<=7;i++)
{
switch (i)
{
case 1: /*process data*/ break;
case 2: /*process data*/ break;
case 3: /*process data*/ break;
case 4: /*process data*/ break;
case 5: /*process data*/ break;
// case 6: // REL 2
case 7: /*process data*/ break;
default: break;
}
}
Can I actually check by adding if statement between those cases? Any better idea?
#define which release is this and then use #ifdef.
switch (i)
{
case 1: /*process data*/ break;
case 2: /*process data*/ break;
case 3: /*process data*/ break;
case 4: /*process data*/ break;
case 5: /*process data*/ break;
#ifdef REL_2
case 6: /*process data*/ break; // <-- executed only for REL_2
#endif
case 7: /*process data*/ break;
default: break;
}
In production code if I have an unexpected result like this I often try to capture and log it.
Throwing an exception may be ok depending on how it's handled. For future-proofing it's usually good to design stuff like this to gracefully handle new values.
Instead of using a 'for' loop, use a 'while' loop with a release number generator function.
Pseudo code:
typedef enum {rel1, rel2} rel_t;
// Where 'ReleaseSequence' is a generator functor. Class that takes rel_t
// in constructor and creates an appropriate functor that returns the
// required sequence of numbers for a particular release
ReleaseSequence seq(rel1);
while (i = seq()) {
switch (i) {
case 1:...
case 2:...
case 3:...
case 4:...
case 5:...
case 6:...
case 7:...
default:
break;
}
}
I find it very strange to see a switch nested within a loop...
It looks like you are executing a pipe. Then why don't you simply define is as one.
For example, a simple array of pointer to functions would do nicely. You can define one such pipe per release, and provide "noop" functions for the parameters to ignore.
If you want to catch unexpected values, shouldn't you put that check in the default part?
...
default:
assert(false); // We should never get here!
Related
Is it possible to obtain switch statements formatted as:
switch (index) {
case 0: /* statement */ break;
case 1: /* statement */ break;
default: break;
}
with Clang-Format?
Yes, you need to set AllowShortCaseLabelsOnASingleLine to true. You can experiment in real-time with your .clang-format configuration on this great website: "clang-format-configurator".
I wrote this part of code and a series of error are being shown. The above mentioned error being the first. What is wrong in the code?
void direction(char ch)
{
switch(ch)
case 'w': if(dirn!=3){dirn=1;}
break;
case 'a': if(dirn!=2){dirn=4;}
break;
case 's': if(dirn!=1){dirn=3;}
break;
case 'd': if(dirn!=4){dirn=2;}
break;
You can choose to omit the opening and closing braces for a switch statement only when you have only one case in your switch block as shown below:
void direction(char ch)
{
switch(ch)
case 'w': if(dirn!=3){dirn=1;}
}
But, if you have got multiple cases to deal with like in your case then you must enclose them inside a pair of opening and closing braces to create a code block for the switch statement as shown below:
void direction(char ch)
{
switch(ch)
{//opening brace for starting of statement block
case 'w': if(dirn!=3){dirn=1;}
break;
case 'a': if(dirn!=2){dirn=4;}
break;
case 's': if(dirn!=1){dirn=3;}
break;
case 'd': if(dirn!=4){dirn=2;}
break;
}//closing brace for closing of statement block
So you will have to either remove all the cases but one OR add the pair of braces to create statement block. In all other cases your code won't compile successfully.
You have forgotten the switch braces :
void direction(char ch)
{
switch(ch)
{
case 'w': if(dirn!=3){dirn=1;}
break;
case 'a': if(dirn!=2){dirn=4;}
break;
case 's': if(dirn!=1){dirn=3;}
break;
case 'd': if(dirn!=4){dirn=2;}
break;
}
}
switch statement requires braces block, where all labels including default one should be:
switch(ch)
{
case 'w': if(dirn!=3) dirn=1;
break;
case 'a': if(dirn!=2) dirn=4;
break;
case 's': if(dirn!=1) dirn=3;
break;
case 'd': if(dirn!=4) dirn=2;
break;
default:
break;
}
The statement after switch must be a compound statement to contain case, default and break. Break got a special meaning here, different from loops. If brace was omitted only next line after switch is part of its statement.
#include<bits/stdc++.h>
using namespace std;
void show(int errorCause)
{
switch(errorCause)
{
case 1:
{
cout<<"in 1\n";
break;
}
case 2: break;
case 3:
{
cout<<"in 3\n";
break;
case 4:
{
cout<<"in 4\n";
case 5: cout<<"in 5\n";
break;
}
}
break;
default:
{
cout<<"in deafult\n";
break;
}
}
return;
}
int main()
{
show(5);
return 0;
}
I used this sample of code and I could not figure out its flow.According to me it should match the default condition as the errorCause does not match anything,but its output is:
in 5
I don't understand why it is not going to default condition?
Here is my build environment details:
compiler:
g++ version 4.8.4 (Ubuntu 4.8.4-2ubuntu1~14.04.3)
System:
Ubuntu 14.04(64-bit)
You pass 5, why should the switch statement not go into 'case 5'?
To make it clear: Remove all these curly braces inside the switch-block, none of them is necessary. The re-align and format the code, then it should be clear.
case/default labels for a switch statement may appear anywhere within that switch statement, except within a nested switch statement.
A famous example of this usage is Duff's device for unrolling loops:
void copy(unsigned char *to, const unsigned char *from, size_t count)
{
size_t n;
if (!count)
return;
n = (count + 7) / 8;
switch (count % 8) {
case 0:
do {
*to++ = *from++;
case 1:
*to++ = *from++;
case 2:
*to++ = *from++;
case 3:
*to++ = *from++;
case 4:
*to++ = *from++;
case 5:
*to++ = *from++;
case 6:
*to++ = *from++;
case 7:
*to++ = *from++;
case 1:
} while (--n > 0);
}
}
(adapted from the original).
At first glance, that doesn't make any sense (and it is somewhat redundant if you allow the compiler to unroll loops for you), but it illustrates that case labels can be placed more or less where you like within the switch statement.
First, don't write code like that. <g>
Second, the reason that it gets to case 5: is simply that there's a case 5: inside the switch statement. It doesn't matter that it's nested inside two levels of curly braces; it's just a label for the code to jump to. It doesn't have to be at the outer level of the switch statement.
It's because actually the switch statement evaluation is "relaxed", so the braces do not matter there. Only case matters, but you can jump right into the middle of a scope by the case (or even to the middle of a loop, see Duff's device).
because the value you passed is 5 , which exactly matches with the switch case parameter.
case 5: cout<<"in 5\n";
break;
if you want to get the default statement then modify the main function as shown below :
int main()
{
show(6);
return 0;
}
hope this helps.
What is wrong with this code:
switch (n)
{
case 0: strcpy(resultString, "Zero");
case 1: strcpy(resultString, "One");
case 2: strcpy(resultString, "Two");
case 3: strcpy(resultString, "Three");
case 4: strcpy(resultString, "Four");
case 5: strcpy(resultString, "Five");
case 6: strcpy(resultString, "Six");
case 7: strcpy(resultString, "Seven");
case 8: strcpy(resultString, "Eight");
case 9: strcpy(resultString, "Nine");
}
printf("%s", resultString);
It always prints "Nine" no matter the value of n. What am I doing wrong??
You need a break statement at the end of each case. Otherwise control falls straight through to the next case.
Change your code to:
switch (n)
{
case 0: strcpy(resultString, "Zero");
break;
case 1: strcpy(resultString, "One");
break;
case 2: strcpy(resultString, "Two");
break;
case 3: strcpy(resultString, "Three");
break;
case 4: strcpy(resultString, "Four");
break;
case 5: strcpy(resultString, "Five");
break;
case 6: strcpy(resultString, "Six");
break;
case 7: strcpy(resultString, "Seven");
break;
case 8: strcpy(resultString, "Eight");
break;
case 9: strcpy(resultString, "Nine");
break;
}
printf("%s", resultString);
You can find the switch statement documented here or in any book on the C language.
You need to break after each case.
case 0:
do soemthing;
break;
case 1:
do something;
break;
In many managed languages, it won't let "one case fall through to another," and throws an error. But C loves to let you do whatever you want!
You missed break; after each case
Example :
case 0: strcpy(resultString, "Zero");break;
..
..
case 8: .... ; break;
..
From the standard :
6.4.2 The switch statement [stmt.switch]
case and default labels in themselves do not alter the flow of control, which continues unimpeded across such labels. To exit from a switch, see break (6.6.1).
6.6.1 The break statement [stmt.break]
The break statement shall occur only in an iteration-statement or a switch statement and causes termination of the smallest enclosing iteration-statement or switch statement; control passes to the statement following the terminated statement, if any.
That means that is you don't use break after each case, you program will enter in the first case who matches the condition and will continue executing every line of the switch until the end.
You should just do something like :
switch( n )
{
case 0:
// ...
break; // <- Note the break
//...
default:
// ...
}
switch(choice)
{
case 1:
uinstance1.addNewProduct(data);
break;
case 2:
break;
case 3:
break;
case 4:
break;
case 5:
break;
case 6:
break;
case 7:
uinstance1.listAllProducts(data);
break;
case 8:
break;
case 9:
break;
case 10:
//name,category,barcode,price,manufacturer,noinstock,soldpermonth,expirydate,discount
// Perishable(string,string,string,double,string,int,int);
Perishable item0("Ferrari","Automobile","9999",2999.99,"Popular",5,0);
data.addNew(item0);
break;
default:
cout<<"Wrong Choice "<<endl;
system("pause");
break;
}
}
Hi ,i have been thinking about this error for quite some time and cant seem to figure out the issue.
error C2361: initialization of 'item0' is skipped by 'default' label
: see declaration of 'item0'
Some help would be appreciated.
Thanks
The whole select block counts as one scope, if you decalare a variable in that scope you need to initialize it in every case statement (every possible execution path). You can avoid it by creating a additional scope in your case to avoid the problem (see the brackets):
switch(choice)
{
case 1:
uinstance1.addNewProduct(data);
break;
case 2:
break;
case 3:
break;
case 4:
break;
case 5:
break;
case 6:
break;
case 7:
uinstance1.listAllProducts(data);
break;
case 8:
break;
case 9:
break;
case 10:
{
//name,category,barcode,price,manufacturer,noinstock,soldpermonth,expirydate,discount
// Perishable(string,string,string,double,string,int,int);
Perishable item0("Ferrari","Automobile","9999",2999.99,"Popular",5,0);
data.addNew(item0);
}
break;
default:
cout<<"Wrong Choice "<<endl;
system("pause");
break;
}
}
MSDN explains the error C2361 aptly:
The initialization of identifier can be skipped in a switch statement. You cannot jump past a declaration with an initializer unless the declaration is enclosed in a block. (Unless it is declared within a block, the variable is within scope until the end of the switch statement.)
Always pay attention to the error numbers they provide vital information about why the error.
You forgot the braces in one of the cases.
case 10:
{
^^^
Perishable item0;
data.addNew(item0);
break;
}
^^^
Your label is crossing initialization which is illegal. Pretty sure moving default to the top should fix it. You can also add code blocks { } around your relevant code. If you still have problems then move your object outside of the switch block.
You can't create variables inside case statements if you don't define explicity the scope.
There is another discussion about that: Variables inside case statement
case 10:
{ // <<-- This gives explicit scope for the stack variable and let's you get rid of the error
Perishable item0;
// ...
}
break;