473,322 Members | 1,781 Online
Bytes | Software Development & Data Engineering Community
Post Job

Home Posts Topics Members FAQ

Join Bytes to post your question to a community of 473,322 software developers and data experts.

error: jump to case label

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
Jul 22 '05 #1
11 17201
Neil Zanella wrote:
I get this error when switching two case labels together with their bodies.


Explain that further or better yet post the offending code.
Jul 22 '05 #2
On Fri, 02 Apr 2004 15:09:00 -0800, Neil Zanella wrote:
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

Jul 22 '05 #3
On Fri, 02 Apr 2004 15:09:00 -0800, Neil Zanella wrote:
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

Jul 22 '05 #4
Bill Seurer <se****@us.ibm.com> wrote in message news:<c4***********@news.rchland.ibm.com>...
Neil Zanella wrote:
I get this error when switching two case labels together with their bodies.


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);
}
Jul 22 '05 #5
Bill Seurer <se****@us.ibm.com> wrote in message news:<c4***********@news.rchland.ibm.com>...
Neil Zanella wrote:
I get this error when switching two case labels together with their bodies.


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);
}
Jul 22 '05 #6
Neil Zanella wrote:
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.
Jul 22 '05 #7
Neil Zanella wrote:
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.
Jul 22 '05 #8
Jacek Dziedzic <ja*************@janowo.net> wrote in message:
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
#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.

Jul 22 '05 #9
Jacek Dziedzic <ja*************@janowo.net> wrote in message:
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
#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.

Jul 22 '05 #10
Neil Zanella wrote:
Jacek Dziedzic <ja*************@janowo.net> wrote in message:
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.


Please don't top-post.
#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);
}


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
}

--
Regards,
Buster.
Jul 22 '05 #11
Neil Zanella wrote:
Jacek Dziedzic <ja*************@janowo.net> wrote in message:
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.


Please don't top-post.
#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);
}


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
}

--
Regards,
Buster.
Jul 22 '05 #12

This thread has been closed and replies have been disabled. Please start a new discussion.

Similar topics

2
by: Andreas | last post by:
Hi, can someone tell me how i do create a jump table in c or c++ ? I tried this: void f() { void *jump_table = { label1,
10
by: Neil Zanella | last post by:
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...
11
by: Steve Hoyer | last post by:
I am trying to deploy my first asp.net app to our webserver (2K server, IIS 5) My start page comes up and you can get to the subsequent pages that are tied into our sql server (2K). Each page...
33
by: Anthony England | last post by:
I am considering general error handling routines and have written a sample function to look up an ID in a table. The function returns True if it can find the ID and create a recordset based on...
2
by: dc15 | last post by:
I am trying to write an application, taking input from various text boxes and appending it to a text file.... it was working, then I had to create various sub procedures to reflect limits set in...
9
by: sam_cit | last post by:
Hi Everyone, I wanted to know as to how a switch case like the following one is converted into a jump table by the compiler, assuming that it does, switch(i) { case 4 : { ... printf("4");
3
by: AJ32 | last post by:
I have written a Dev-C++ program that needs to read and write to and from files, also it needs to use the goto function to jump to different parts of the program, after I finished writting the...
12
by: Franz Hose | last post by:
the following program, when compiled with gcc and '-std=c99', gcc says test.c:6: error: jump into scope of identifier with variably modified type that is, it does not even compile. ...
1
by: KRISHNA PRAVI | last post by:
the error is "runtime error object expected" here is the code....................................................................................... <script language="javascript"...
0
by: DolphinDB | last post by:
Tired of spending countless mintues downsampling your data? Look no further! In this article, you’ll learn how to efficiently downsample 6.48 billion high-frequency records to 61 million...
0
isladogs
by: isladogs | last post by:
The next Access Europe meeting will be on Wednesday 6 Mar 2024 starting at 18:00 UK time (6PM UTC) and finishing at about 19:15 (7.15PM). In this month's session, we are pleased to welcome back...
1
isladogs
by: isladogs | last post by:
The next Access Europe meeting will be on Wednesday 6 Mar 2024 starting at 18:00 UK time (6PM UTC) and finishing at about 19:15 (7.15PM). In this month's session, we are pleased to welcome back...
0
by: Vimpel783 | last post by:
Hello! Guys, I found this code on the Internet, but I need to modify it a little. It works well, the problem is this: Data is sent from only one cell, in this case B5, but it is necessary that data...
0
by: jfyes | last post by:
As a hardware engineer, after seeing that CEIWEI recently released a new tool for Modbus RTU Over TCP/UDP filtering and monitoring, I actively went to its official website to take a look. It turned...
0
by: ArrayDB | last post by:
The error message I've encountered is; ERROR:root:Error generating model response: exception: access violation writing 0x0000000000005140, which seems to be indicative of an access violation...
1
by: PapaRatzi | last post by:
Hello, I am teaching myself MS Access forms design and Visual Basic. I've created a table to capture a list of Top 30 singles and forms to capture new entries. The final step is a form (unbound)...
0
by: af34tf | last post by:
Hi Guys, I have a domain whose name is BytesLimited.com, and I want to sell it. Does anyone know about platforms that allow me to list my domain in auction for free. Thank you
0
isladogs
by: isladogs | last post by:
The next Access Europe User Group meeting will be on Wednesday 3 Apr 2024 starting at 18:00 UK time (6PM UTC+1) and finishing by 19:30 (7.30PM). In this session, we are pleased to welcome former...

By using Bytes.com and it's services, you agree to our Privacy Policy and Terms of Use.

To disable or enable advertisements and analytics tracking please visit the manage ads & tracking page.