error: jump to case label

[Neil Zanella]

Hello,

Does anyone know what the following g++ compiler error message means:

error: jump to case label

I get this error when switching two case labels together with their bodies.
I have no setjmp/longjmp or gotos in my program.

Thanks,

Neil

 

[Bill Seurer]

I get this error when switching two case labels together with their bodies.

Explain that further or better yet post the offending code.

 

[James Gregory]

Hello,

Does anyone know what the following g++ compiler error message means:

error: jump to case label

I get this error when switching two case labels together with their bodies.
I have no setjmp/longjmp or gotos in my program.

Perhaps the problem is "jump to case label croses initialization"?

The following is not allowed:

switch (a)
{
case 1:
int a = 6;
//stuff
break;

case 2:
//stuff
break;
}

The following is allowed:

switch (a)
{
case 1:
{
int a = 6;
//stuff
}
break;

case 2:
//stuff
break;
}

James

 

[Neil Zanella]

Explain that further or better yet post the offending code.

Sure I will. Here is the code. Uncommenting the lines for case 1 produces
the compiler error message. Furthermore, out of curiosity, as an unrelated
matter, I am quite interested in knowing how come the program starts looping
when some large number is entered.

Thanks,

Neil

#include <iostream>

int main() {
unsigned int x; do {
std::cout << "Please enter an integer: " << std::flush;
std::cin >> x;
switch (x) {
case 0:
std::cout << "Hello!" << std::endl;
break;
default:
unsigned int y = ++x;
std::cout << "You could have entered " << y;
std::cout << ". Why didn't you?" << std::endl;
break;
//case 1:
// std::cout << "What??? You entered one?" << std::endl;
// break;
}
} while (x != 0);
}

 

[Jacek Dziedzic]

Sure I will. Here is the code. Uncommenting the lines for case 1 produces
the compiler error message.

Try moving them before the 'default'

Furthermore, out of curiosity, as an unrelated
matter, I am quite interested in knowing how come the program starts looping
when some large number is entered.

cin.fail() is set and all subsequent ">>" operations are ignored,
with x unmodified every time. Hence, if x happened to be non-zero,
the loop iterates infinitely.

Thanks,

Neil

#include <iostream>

int main() {
unsigned int x; do {
std::cout << "Please enter an integer: " << std::flush;
std::cin >> x;
switch (x) {
case 0:
std::cout << "Hello!" << std::endl;
break;
default:
unsigned int y = ++x;
std::cout << "You could have entered " << y;
std::cout << ". Why didn't you?" << std::endl;
break;
//case 1:
// std::cout << "What??? You entered one?" << std::endl;
// break;
}
} while (x != 0);
}

HTH,
- J.

 

[Neil Zanella]

Try moving them before the 'default'

Thank you for your reply...
I know that works but that doens't really explain the nature of the problem.

Regards,

Neil

 

[Buster]

Thank you for your reply...
I know that works but that doens't really explain the nature of the problem.

Please don't top-post.

James Gregory is right. The 'jump' in the error message is the computed
goto effected by the switch statement. When x = 1, the switch acts like
this:

goto case_label_1;
// ...
unsigned int y = ++ x;
// ...
case_label_1:
// here y is uninitialized

The problem is that the initialization of y is skipped when x == 1.
When the "case 1:" label is reached, stack space has been allocated for
y but its value has not been initialized. This is not allowed.

The general way round this situation is to make the scope of y smaller
by adding braces:

switch (x)
{
default:
unsigned z; // this is OK since z is uninitialized
{
unsigned y = x + 1;
// ...
}
case 1:
z = 2;
// ...
// cannot refer to y here so no problem with y's initialization
}