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 doe

Answer 1

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 
#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\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.