digitalmars.D.learn - switch(string)

David Ferenczi (24/24) Jun 18 2008 The following code gives the following compilation error:

Robert Fraser (7/39) Jun 18 2008 BAR1 and BAR2 are not constant. In D1, try:

David Ferenczi (6/46) Jun 18 2008 Thank you very much for the quick reply. Does it mean that a

Robert Fraser (11/56) Jun 18 2008 Yes, but you can write it as:

David Ferenczi (2/66) Jun 18 2008 Thank you very much! :-) I'm on the way to clarity. ;-)
Nick Sabalausky (7/9) Jun 18 2008 To clarify:

BCS (4/21) Jun 18 2008 Almost (the difference is almost irrelevant, I think), auto is a do-noth...

David Ferenczi <raggae ferenczi.net> writes:

The following code gives the following compilation error:

Error: case must be a string or an integral constant, not BAR1
Error: case must be a string or an integral constant, not BAR2

--------------------------8<------------------------------------
int main(string[] args)
{
    string BAR1 = "bar1";
    string BAR2 = "bar2";
    string myString = "bar3";

    switch (myString)
    {
        case BAR1:
            break;

        case BAR2:
            break;

        defualt:
            break;
    }

    return 0;
}
--------------------------8<------------------------------------

Can somebody explain me why?

Thanks,
David

Jun 18 2008

Robert Fraser <fraserofthenight gmail.com> writes:

David Ferenczi wrote:
 The following code gives the following compilation error:
 
 Error: case must be a string or an integral constant, not BAR1
 Error: case must be a string or an integral constant, not BAR2
 
 --------------------------8<------------------------------------
 int main(string[] args)
 {
     string BAR1 = "bar1";
     string BAR2 = "bar2";
     string myString = "bar3";
 
     switch (myString)
     {
         case BAR1:
             break;
 
         case BAR2:
             break;
 
         defualt:
             break;
     }
 
     return 0;
 }
 --------------------------8<------------------------------------
 
 Can somebody explain me why?
 
 Thanks,
 David

BAR1 and BAR2 are not constant. In D1, try:
     const BAR1 = "bar1";
In D2, you can also try:
     invariant BAR1 = "bar1";
     enum BAR1 = "bar1";
The enum version and the const version in D1 will not take up any storage.

Jun 18 2008

David Ferenczi <raggae ferenczi.net> writes:

Robert Fraser wrote:

 David Ferenczi wrote:
 The following code gives the following compilation error:
 
 Error: case must be a string or an integral constant, not BAR1
 Error: case must be a string or an integral constant, not BAR2
 
 --------------------------8<------------------------------------
 int main(string[] args)
 {
     string BAR1 = "bar1";
     string BAR2 = "bar2";
     string myString = "bar3";
 
     switch (myString)
     {
         case BAR1:
             break;
 
         case BAR2:
             break;
 
         defualt:
             break;
     }
 
     return 0;
 }
 --------------------------8<------------------------------------
 
 Can somebody explain me why?
 
 Thanks,
 David

 
 BAR1 and BAR2 are not constant. In D1, try:
      const BAR1 = "bar1";
 In D2, you can also try:
      invariant BAR1 = "bar1";
      enum BAR1 = "bar1";
 The enum version and the const version in D1 will not take up any storage.

Thank you very much for the quick reply. Does it mean that a
constant/invariant doesn't need an explicit storage type?

I thought that string was invariant, so it was constant. Isn't it right?

Thanks,
David

Jun 18 2008

Robert Fraser <fraserofthenight gmail.com> writes:

David Ferenczi wrote:
 Robert Fraser wrote:
 
 David Ferenczi wrote:
 The following code gives the following compilation error:

 Error: case must be a string or an integral constant, not BAR1
 Error: case must be a string or an integral constant, not BAR2

 --------------------------8<------------------------------------
 int main(string[] args)
 {
     string BAR1 = "bar1";
     string BAR2 = "bar2";
     string myString = "bar3";

     switch (myString)
     {
         case BAR1:
             break;

         case BAR2:
             break;

         defualt:
             break;
     }

     return 0;
 }
 --------------------------8<------------------------------------

 Can somebody explain me why?

 Thanks,
 David

 BAR1 and BAR2 are not constant. In D1, try:
      const BAR1 = "bar1";
 In D2, you can also try:
      invariant BAR1 = "bar1";
      enum BAR1 = "bar1";
 The enum version and the const version in D1 will not take up any storage.

 
 Thank you very much for the quick reply. Does it mean that a
 constant/invariant doesn't need an explicit storage type?

Yes, but you can write it as:

invariant string BAR1 = "bar1";

If "string" isn't specified it will be deduced.

 I thought that string was invariant, so it was constant. Isn't it right?

string is invariant(char)[]. So the letters are constant, but not the 
reference. This means:

string x = "abc";
x[2] ='d'; // ERROR
x = "abd"; // Okay

It's kind of confusing at first, and I have yet to be convinced of its 
usefulness, but there ya go.

Jun 18 2008

David Ferenczi <raggae ferenczi.net> writes:

Robert Fraser wrote:

 David Ferenczi wrote:
 Robert Fraser wrote:
 
 David Ferenczi wrote:
 The following code gives the following compilation error:

 Error: case must be a string or an integral constant, not BAR1
 Error: case must be a string or an integral constant, not BAR2

 --------------------------8<------------------------------------
 int main(string[] args)
 {
     string BAR1 = "bar1";
     string BAR2 = "bar2";
     string myString = "bar3";

     switch (myString)
     {
         case BAR1:
             break;

         case BAR2:
             break;

         defualt:
             break;
     }

     return 0;
 }
 --------------------------8<------------------------------------

 Can somebody explain me why?

 Thanks,
 David

 BAR1 and BAR2 are not constant. In D1, try:
      const BAR1 = "bar1";
 In D2, you can also try:
      invariant BAR1 = "bar1";
      enum BAR1 = "bar1";
 The enum version and the const version in D1 will not take up any
 storage.

 
 Thank you very much for the quick reply. Does it mean that a
 constant/invariant doesn't need an explicit storage type?

 
 Yes, but you can write it as:
 
 invariant string BAR1 = "bar1";
 
 If "string" isn't specified it will be deduced.
 
 I thought that string was invariant, so it was constant. Isn't it right?

 
 string is invariant(char)[]. So the letters are constant, but not the
 reference. This means:
 
 string x = "abc";
 x[2] ='d'; // ERROR
 x = "abd"; // Okay
 
 It's kind of confusing at first, and I have yet to be convinced of its
 usefulness, but there ya go.

Thank you very much! :-) I'm on the way to clarity. ;-)

Jun 18 2008

"Nick Sabalausky" <a a.a> writes:

"Robert Fraser" <fraserofthenight gmail.com> wrote in message 
news:g3al00$20ml$1 digitalmars.com...
 invariant string BAR1 = "bar1";

 If "string" isn't specified it will be deduced.

To clarify:

invariant BAR1 = "bar1";

Is shorthand for:

invariant auto BAR1 = "bar1";

And "auto" means the type is deduced.

Jun 18 2008

BCS <ao pathlink.com> writes:

Reply to Nick,

 "Robert Fraser" <fraserofthenight gmail.com> wrote in message
 news:g3al00$20ml$1 digitalmars.com...
 
 invariant string BAR1 = "bar1";
 
 If "string" isn't specified it will be deduced.
 

 To clarify:
 
 invariant BAR1 = "bar1";
 
 Is shorthand for:
 
 invariant auto BAR1 = "bar1";
 
 And "auto" means the type is deduced.
 

Almost (the difference is almost irrelevant, I think), auto is a do-nothing 
attribute that is used if no other attributes are used. The syntax for type 
deduction is "<attributes> <name> = <exp>"

Jun 18 2008

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digitalmars.D.learn - switch(string)