Error: Jump to case label in switch statement

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

Ainu

What code are you trying to compile? What compiler are you using? Have you enclosed each case block in braces? – Plastometer

that's one impressively roundabout error message – Xantha

Indeed, you absolutely could jump to the wrong conclusion from this! – Bowsprit

Answer 1

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 << i;
}

Vivianviviana

See this fixed LLVM bug report for other explanations: llvm.org/bugs/show_bug.cgi?id=7789 – Corina

Is goto good coding practice in this situation? – Respectful

Answer 2

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;
}    

Essayistic

cleaner fix instruction – Coady

Will there be any problem if I put the break statement inside the curly braces? – Varga

I always add braces, all sorts of weird scoping problems can happen if you do not. @VishalSharma its fine to have the break statement inside the braces – Sven

I had never thought to such a simple solution, thank you! +1 – Mannino

Answer 3

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

Digged

Answer 4

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 包子露宪 六四事件 法轮功 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 <cstdio>
#else
#include <stdio.h>
#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\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.

Ileenileitis