Recommendation for the error: jump to case label in switch statement in c++ error

Question

I get the “Error: Jump to case label in switch statement in C++” error as the title says. How can I fix it so the error goes away? Here is my detail:

#include <iostream>
 int main() 
 {
  int choice;
  std::cin >> choice;
  switch(choice)
  {
  case 1:
  int i=0;
  break;
  case 2: // error here 
  }
 }

When I operated it, I received the error text:

Error: Jump to case label.

I appreciate any help from you.

lyytutoria · Accepted Answer

The cause:

This problem because declaration of variables in case will be shown in subsequent case unless HTML13 blocks are used. However, they won’t be initialized because the initialization code belongs to another case.

Solution:

You can check whether foo equals 1 by the following code. If it equals 1, all is fine but if it equals 2, then we’ll accidentally use i, it exists but contains garbage.

switch(foo) {
case 1:
int i = 42; // i shows all the ways to the end of the switch which are existing 
dostuff(i);
break;
case 2:
dostuff(i*2); // i also is in scope here, but it is not initialized!
}

You can solve this probblem by wrapping the case in an explicit block:

switch(foo) {
 case 1:
 {
 int i = 42; // only within the { }, i can exist
 dostuff(i);
 break;
 }
 case 2:
 dostuff(123); // i can not used accidentally now
}

Moreover, you also can describe a more fancy version of goto by using switch statements. Below is an analogous piece of code that has the same issue, but it uses a goto rather than a switch.

int main() {
if(rand() % 2) // Toss a coin
goto end;

int i = 42;

end:
// We either can skip the declaration of i or not,
// but one of the two ways the variable i exists here, as
// variable scopes are resolved at compile time.
// Whether the *initialization* code is runned or not, eventhough,
// it depends on whether rand returned 0 or 1.
std::cout << i;
}

Adam Dumas · Answer

Problems are caused by the declaration of new variables within case statements. This problem can be solved by enclosing all case {} statements.

switch(choice)
{
 case 1: {
 // .......
 }break;
 case 2: {
 // .......
 }break;
 case 3: {
 // .......
 }break;
}

James Singh · Answer

C++11 standard for jumping over initializations

JohannesD provided an explanation, now we will discuss the standards.

The C++11 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 ]

GCC 5.2 now has an error message that reads:

crosses initialization of

C

C allows it: goto past initialization

The C99 Annex I standard draft Annex I "Common Warnings" states:

2 A block with initialization of an object that has automatic storage duration is jumped into

Della Thomas · Answer

JohannesD's answer was correct, but it doesn't address a specific aspect of the problem.

He gives an example where he declares and initializes i in case 1 and then attempts to use it in the second case. He argues that i could be used if it was switched to case 2 without being initialized. This is why there is a compilation error. One could argue that variables declared in a particular case would not be used in any other cases. Take this example:

switch(choice) {
 case 1:
 int i = 10; // i is never used outside of this case
 printf("i = %d
", i);
 break;
 case 2:
 int j = 20; // j is never used outside of this case
 printf("j = %d
", j);
 break;
}

This program could be expected to compile since i is used only in cases where j is declared. It doesn't work in C++. Ciro Zi Santilli Bai Zi Liu Si Shi Jian FA Lun Gong explained. We can't skip to case 2: because that would skip the initialization of i. And even though case 2 does not use i at any time, this is still prohibited in C++.

This program can be compile as C with some modifications (a #ifdef to #include and a semicolon following the labels, since labels cannot be followed by statements and declarations in C).

// Disable warning issued by MSVC about scanf being deprecated
#ifdef _MSC_VER
#define _CRT_SECURE_NO_WARNINGS
#endif

#ifdef __cplusplus
#include 
#else
#include 
#endif

int main() {

 int choice;
 printf("Please enter 1 or 2: ");
 scanf("%d", &choice);

 switch(choice) {
 case 1:
 ;
 int i = 10; // i is never used outside of this case
 printf("i = %d
", i);
 break;
 case 2:
 ;
 int j = 20; // j is never used outside of this case
 printf("j = %d
", j);
 break;
 }
}

An online compiler such as http://rextester.com allows you to quickly compile it in C or C++ using MSVC, GCC, or Clang. It works fine in C (just set STDIN). C++ does not accept it.

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Recommendation for the error: jump to case label in switch statement in c++ error