Error Jump to case label

Question

I wrote a program which involves use of switch statements    However on compilation it shows      Error  Jump to case label    Why does it do that    include  lt iostream gt   include  lt cstdlib gt   include  lt fstream gt   include  lt string gt   using namespace std   class contact   public      string name      int phonenumber      string address      contact             name   Noname           phonenumber  0          address   Noaddress             int main         contact  d      d   new contact 200       string name add      int choice modchoice t   Variable for switch statement     int phno phno1      int i 0      int initsize 0  i1 0   i is declared as a static int variable     bool flag false flag no blank false         TAKE DATA FROM FILES            We create 3 files names  phone numbers  Address and then abstract the data from these files first      fstream f1      fstream f2      fstream f3      string file input name      string file input address      int file input number       f1 open    names        while f1 gt  gt file input name           d i  name file input name          i              initsize i       f2 open    numbers        while f2 gt  gt file input number           d i1  phonenumber file input number          i1              i1 0       f3 open    address        while f3 gt  gt file input address           d i1  address file input address          i1               cout lt  lt   tWelcome to the phone Directory n    Welcome Message     do            do-While Loop Starts         cout lt  lt  Select   n1 Add New Contact n2 Update Existing Contact n3 Display All Contacts n4 Search for a Contact n5 Delete a  Contact n6 Exit PhoneBook n n n    Display all options         cin gt  gt choice   Input Choice from user          switch choice    Switch Loop Starts         case 1              i     increment i so that values are now taken from the program and stored as different variables             i1                do                  cout lt  lt   nEnter The Name n                   cin gt  gt name                  if name       cout lt  lt  Blank Entries are not allowed                   flag no blank true                                 while flag no blank  true               flag no blank false              d i  name name              cout lt  lt   nEnter the Phone Number n               cin gt  gt phno              d i1  phonenumber phno              cout lt  lt   nEnter the address n               cin gt  gt add              d i1  address add              i1                i                break   Exit Case 1 to the main menu         case 2              cout lt  lt   nEnter the name n    Here it is assumed that no two contacts can have same contact number or address but may have the same name              cin gt  gt name              int k 0 val              cout lt  lt   n nSearching          n n               for int j 0 j lt  i j                     if d j  name  name                       k                        cout lt  lt k lt  lt    t  lt  lt d j  name lt  lt   t  lt  lt d j  phonenumber lt  lt   t  lt  lt d j  address lt  lt   n n                       val j                                              char ch              cout lt  lt   nTotal of   lt  lt k lt  lt   Entries were found    Do you wish to edit  n               string staticname              staticname d val  name              cin gt  gt ch              if ch   y    ch   Y                    cout lt  lt  Which entry do you wish to modify   enter the old telephone number  n                   cin gt  gt phno                  for int j 0 j lt  i j                         if d j  phonenumber  phno  amp  amp  staticname  d j  name                           cout lt  lt  Do you wish to change the name  n                           cin gt  gt ch                          if ch   y   ch   Y                                cout lt  lt  Enter new name n                               cin gt  gt name                              d j  name name                                                    cout lt  lt  Do you wish to change the number  n                           cin gt  gt ch                          if ch   y   ch   Y                                cout lt  lt  Enter the new number n                               cin gt  gt phno1                              d j  phonenumber phno1                                                    cout lt  lt  Do you wish to change the address  n                           cin gt  gt ch                          if ch   y   ch   Y                                cout lt  lt  Enter the new address n                               cin gt  gt add                              d j  address add                                                                                              break          case 3                 cout lt  lt   n tContents of PhoneBook  n n tNames tPhone-Numbers tAddresses               for int t 0 t lt  i t                     cout lt  lt t 1 lt  lt    t  lt  lt d t  name lt  lt   t  lt  lt d t  phonenumber lt  lt   t  lt  lt d t  address                            break                                         while flag  false       return 0

User · Answer

JohannesD s answer is correct  but I feel it isn t entirely clear on an aspect of the problem   The example he gives declares and initializes the variable i in case 1  and then tries to use it in case 2  His argument is that if the switch went straight to case 2  i would be used without being initialized  and this is why there s a compilation error  At this point  one could think that there would be no problem if variables declared in a case were never used in other cases  For example   switch choice        case 1          int i   10     i is never used outside of this case         printf  i    d n   i           break      case 2          int j   20     j is never used outside of this case         printf  j    d n   j           break      One could expect this program to compile  since both i and j are used only inside the cases that declare them  Unfortunately  in C   it doesn t compile  as Ciro Santilli               explained  we simply can t jump to case 2   because this would skip the declaration with initialization of i  and even though case 2 doesn t use i at all  this is still forbidden in C     Interestingly  with some adjustments  an  ifdef to  include the appropriate header  and a semicolon after the labels  because labels can only be followed by statements  and declarations do not count as statements in C   this program does compile as C      Disable warning issued by MSVC about scanf being deprecated  ifdef  MSC VER  define  CRT SECURE NO WARNINGS  endif   ifdef   cplusplus  include  lt cstdio gt   else  include  lt stdio h gt   endif  int main          int choice      printf  Please enter 1 or 2          scanf   d    amp choice        switch choice            case 1                            int i   10     i is never used outside of this case             printf  i    d n   i               break          case 2                            int j   20     j is never used outside of this case             printf  j    d n   j               break            Thanks to an online compiler like http   rextester com you can quickly try to compile it either as C or C    using MSVC  GCC or Clang  As C it always works  just remember to set STDIN    as C   no compiler accepts it

User · Answer

C  11 standard on jumping over some initializations  JohannesD gave an explanation  now for the standards   The C  11 N3337 standard draft 6 7  Declaration statement  says      3 It is possible to transfer into a block  but not in a way that bypasses declarations with initialization  A   program that jumps  87  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        87  The transfer from the condition of a switch statement to a case label is considered a jump in this respect         Example   void f                 goto lx        ill-formed  jump into scope of a          ly    X a   1           lx    goto ly        OK  jump implies destructor                  call for a followed by construction                  again immediately following label ly             end example     As of GCC 5 2  the error message now says      crosses initialization of    C  C allows it  c99 goto past initialization  The C99 N1256 standard draft Annex I  Common warnings  says      2 A block with initialization of an object that has automatic storage duration is jumped into

User · Answer

The problem is that variables declared in one case are still visible in the subsequent cases unless an explicit      block is used  but they will not be initialized because the initialization code belongs to another case   In the following code  if foo equals 1  everything is ok  but if it equals 2  we ll accidentally use the i variable which does exist but probably contains garbage   switch foo      case 1      int i   42     i exists all the way to the end of the switch     dostuff i       break    case 2      dostuff i 2      i is  also  in scope here  but is not initialized      Wrapping the case in an explicit block solves the problem   switch foo      case 1                int i   42     i only exists within the              dostuff i           break          case 2      dostuff 123      Now you cannot use i accidentally     Edit  To further elaborate  switch statements are just a particularly fancy kind of a goto  Here s an analoguous piece of code exhibiting the same issue but using a goto instead of a switch   int main         if rand     2     Toss a coin         goto end       int i   42     end         We either skipped the declaration of i or not         but either way the variable i exists here  because        variable scopes are resolved at compile time         Whether the  initialization  code was run  though         depends on whether rand returned 0 or 1      std  cout  lt  lt  i

User · Answer

Declaration of new variables in case statements is what causing problems  Enclosing all case statements in    will limit the scope of newly declared variables to the currently executing case which solves the problem   switch choice        case 1                           break      case 2                           break      case 3                           break

[c++] Error: Jump to case label

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