Spurious warning gcc -Wuninitialized
OS: Debian 9
compiler: gcc 8.2.0 (installed from buster (testing) repository)
I know that using things from debian testing branch is dangerous, but debian testing is usually stable, and gcc 8.2 has been released as stable, so it shouldn't have many bugs.
in this function:
int user_tui (const char *title, const char *subtitle)
{
int action;
// action = USER_IFACE_ACT_FOO;
show_help();
user_tui_show_log(title, subtitle);
action = usr_input();
return action;
}
It is reporting the following error (-Wall -Werror and also -O3 -march=native):
/.../modules//user//src//user_tui.c: In function ‘user_tui’:
/.../modules//user//src//user_tui.c:91:9: error: ‘action’ may be used uninitialized in this function [-Werror=maybe-uninitialized]
return action;
^~~~~~
cc1: all warnings being treated as errors
When I uncomment the initialization, the error is still there. I think it shouldn't even be needed, as there is no conditional or anything that would ever block the assignment action = usr_input();.
Is it a spurious warning, or is it legit?
I would say it is a bug in gcc; it can't even be considered spurious.
gcc gcc-warning
edited Nov 22 '18 at 15:02
harper
10.3k44285
asked Nov 22 '18 at 13:36
Cacahuete FritoCacahuete Frito
349317
add a comment |
OS: Debian 9
compiler: gcc 8.2.0 (installed from buster (testing) repository)
I know that using things from debian testing branch is dangerous, but debian testing is usually stable, and gcc 8.2 has been released as stable, so it shouldn't have many bugs.
in this function:
int user_tui (const char *title, const char *subtitle)
{
int action;
// action = USER_IFACE_ACT_FOO;
show_help();
user_tui_show_log(title, subtitle);
action = usr_input();
return action;
}
It is reporting the following error (-Wall -Werror and also -O3 -march=native):
/.../modules//user//src//user_tui.c: In function ‘user_tui’:
/.../modules//user//src//user_tui.c:91:9: error: ‘action’ may be used uninitialized in this function [-Werror=maybe-uninitialized]
return action;
^~~~~~
cc1: all warnings being treated as errors
When I uncomment the initialization, the error is still there. I think it shouldn't even be needed, as there is no conditional or anything that would ever block the assignment action = usr_input();.
Is it a spurious warning, or is it legit?
I would say it is a bug in gcc; it can't even be considered spurious.
gcc gcc-warning
edited Nov 22 '18 at 15:02
harper
10.3k44285
asked Nov 22 '18 at 13:36
Cacahuete FritoCacahuete Frito
349317
1
Show the definition ofusr_inputfunction.
– Maxim Egorushkin
Nov 22 '18 at 13:46
Really no conditions, also not in usr_input() itself, maybe by using exceptions which running out of the initializing scope? BTW "gcc 8.2 has been released as stable, so it shouldn't have many bugs." is definitly wrong if you look in the bugtracker. There are "some" open bugs, some are open and unfixed for multiple major versions... But you are right, such an easy to see bug I also would not expect.
– Klaus
Nov 22 '18 at 13:49
1
@Maxim You were right, I had to look insideusr_input(). But I would say that the variable returned byusr_input()is the one that is being used uninitialized, and notactionitself.actionis really being initialized, although with garbage in the case of an uninitialized return ofusr_input(). I would call that a bug in gcc.
– Cacahuete Frito
Nov 22 '18 at 14:32
Please don't change the topic. Instead use theAcceptbutton at the answer that solved your problem. That's the way how StackOverflow works. It's also okay toacceptthe own answer.
– harper
Nov 22 '18 at 15:04
Ok. However, as the answer is mine, I can't until 2 days from now.
– Cacahuete Frito
Nov 22 '18 at 15:05
add a comment |
OS: Debian 9
compiler: gcc 8.2.0 (installed from buster (testing) repository)
I know that using things from debian testing branch is dangerous, but debian testing is usually stable, and gcc 8.2 has been released as stable, so it shouldn't have many bugs.
in this function:
int user_tui (const char *title, const char *subtitle)
{
int action;
// action = USER_IFACE_ACT_FOO;
show_help();
user_tui_show_log(title, subtitle);
action = usr_input();
return action;
}
It is reporting the following error (-Wall -Werror and also -O3 -march=native):
/.../modules//user//src//user_tui.c: In function ‘user_tui’:
/.../modules//user//src//user_tui.c:91:9: error: ‘action’ may be used uninitialized in this function [-Werror=maybe-uninitialized]
return action;
^~~~~~
cc1: all warnings being treated as errors
When I uncomment the initialization, the error is still there. I think it shouldn't even be needed, as there is no conditional or anything that would ever block the assignment action = usr_input();.
Is it a spurious warning, or is it legit?
I would say it is a bug in gcc; it can't even be considered spurious.
gcc gcc-warning
edited Nov 22 '18 at 15:02
harper
10.3k44285
asked Nov 22 '18 at 13:36
Cacahuete FritoCacahuete Frito
349317
OS: Debian 9
compiler: gcc 8.2.0 (installed from buster (testing) repository)
I know that using things from debian testing branch is dangerous, but debian testing is usually stable, and gcc 8.2 has been released as stable, so it shouldn't have many bugs.
in this function:
int user_tui (const char *title, const char *subtitle)
{
int action;
// action = USER_IFACE_ACT_FOO;
show_help();
user_tui_show_log(title, subtitle);
action = usr_input();
return action;
}
It is reporting the following error (-Wall -Werror and also -O3 -march=native):
/.../modules//user//src//user_tui.c: In function ‘user_tui’:
/.../modules//user//src//user_tui.c:91:9: error: ‘action’ may be used uninitialized in this function [-Werror=maybe-uninitialized]
return action;
^~~~~~
cc1: all warnings being treated as errors
When I uncomment the initialization, the error is still there. I think it shouldn't even be needed, as there is no conditional or anything that would ever block the assignment action = usr_input();.
Is it a spurious warning, or is it legit?
I would say it is a bug in gcc; it can't even be considered spurious.
gcc gcc-warning
gcc gcc-warning
edited Nov 22 '18 at 15:02
harper
10.3k44285
asked Nov 22 '18 at 13:36
Cacahuete FritoCacahuete Frito
349317
edited Nov 22 '18 at 15:02
harper
10.3k44285
asked Nov 22 '18 at 13:36
Cacahuete FritoCacahuete Frito
349317
edited Nov 22 '18 at 15:02
harper
10.3k44285
edited Nov 22 '18 at 15:02
harper
10.3k44285
edited Nov 22 '18 at 15:02
harper
10.3k44285
10.3k44285
asked Nov 22 '18 at 13:36
Cacahuete FritoCacahuete Frito
349317
asked Nov 22 '18 at 13:36
Cacahuete FritoCacahuete Frito
349317
asked Nov 22 '18 at 13:36
Cacahuete FritoCacahuete Frito
349317
349317
1
Show the definition ofusr_inputfunction.
– Maxim Egorushkin
Nov 22 '18 at 13:46
Really no conditions, also not in usr_input() itself, maybe by using exceptions which running out of the initializing scope? BTW "gcc 8.2 has been released as stable, so it shouldn't have many bugs." is definitly wrong if you look in the bugtracker. There are "some" open bugs, some are open and unfixed for multiple major versions... But you are right, such an easy to see bug I also would not expect.
– Klaus
Nov 22 '18 at 13:49
1
@Maxim You were right, I had to look insideusr_input(). But I would say that the variable returned byusr_input()is the one that is being used uninitialized, and notactionitself.actionis really being initialized, although with garbage in the case of an uninitialized return ofusr_input(). I would call that a bug in gcc.
– Cacahuete Frito
Nov 22 '18 at 14:32
Please don't change the topic. Instead use theAcceptbutton at the answer that solved your problem. That's the way how StackOverflow works. It's also okay toacceptthe own answer.
– harper
Nov 22 '18 at 15:04
Ok. However, as the answer is mine, I can't until 2 days from now.
– Cacahuete Frito
Nov 22 '18 at 15:05
add a comment |
1
Show the definition ofusr_inputfunction.
– Maxim Egorushkin
Nov 22 '18 at 13:46
Really no conditions, also not in usr_input() itself, maybe by using exceptions which running out of the initializing scope? BTW "gcc 8.2 has been released as stable, so it shouldn't have many bugs." is definitly wrong if you look in the bugtracker. There are "some" open bugs, some are open and unfixed for multiple major versions... But you are right, such an easy to see bug I also would not expect.
– Klaus
Nov 22 '18 at 13:49
1
@Maxim You were right, I had to look insideusr_input(). But I would say that the variable returned byusr_input()is the one that is being used uninitialized, and notactionitself.actionis really being initialized, although with garbage in the case of an uninitialized return ofusr_input(). I would call that a bug in gcc.
– Cacahuete Frito
Nov 22 '18 at 14:32
Please don't change the topic. Instead use theAcceptbutton at the answer that solved your problem. That's the way how StackOverflow works. It's also okay toacceptthe own answer.
– harper
Nov 22 '18 at 15:04
Ok. However, as the answer is mine, I can't until 2 days from now.
– Cacahuete Frito
Nov 22 '18 at 15:05
1
1
Show the definition of
usr_input function.– Maxim Egorushkin
Nov 22 '18 at 13:46
Show the definition of
usr_input function.– Maxim Egorushkin
Nov 22 '18 at 13:46
Really no conditions, also not in usr_input() itself, maybe by using exceptions which running out of the initializing scope? BTW "gcc 8.2 has been released as stable, so it shouldn't have many bugs." is definitly wrong if you look in the bugtracker. There are "some" open bugs, some are open and unfixed for multiple major versions... But you are right, such an easy to see bug I also would not expect.
– Klaus
Nov 22 '18 at 13:49
Really no conditions, also not in usr_input() itself, maybe by using exceptions which running out of the initializing scope? BTW "gcc 8.2 has been released as stable, so it shouldn't have many bugs." is definitly wrong if you look in the bugtracker. There are "some" open bugs, some are open and unfixed for multiple major versions... But you are right, such an easy to see bug I also would not expect.
– Klaus
Nov 22 '18 at 13:49
1
1
@Maxim You were right, I had to look inside
usr_input(). But I would say that the variable returned by usr_input() is the one that is being used uninitialized, and not action itself. action is really being initialized, although with garbage in the case of an uninitialized return of usr_input(). I would call that a bug in gcc.– Cacahuete Frito
Nov 22 '18 at 14:32
@Maxim You were right, I had to look inside
usr_input(). But I would say that the variable returned by usr_input() is the one that is being used uninitialized, and not action itself. action is really being initialized, although with garbage in the case of an uninitialized return of usr_input(). I would call that a bug in gcc.– Cacahuete Frito
Nov 22 '18 at 14:32
Please don't change the topic. Instead use the
Accept button at the answer that solved your problem. That's the way how StackOverflow works. It's also okay to accept the own answer.– harper
Nov 22 '18 at 15:04
Please don't change the topic. Instead use the
Accept button at the answer that solved your problem. That's the way how StackOverflow works. It's also okay to accept the own answer.– harper
Nov 22 '18 at 15:04
Ok. However, as the answer is mine, I can't until 2 days from now.
– Cacahuete Frito
Nov 22 '18 at 15:05
Ok. However, as the answer is mine, I can't until 2 days from now.
– Cacahuete Frito
Nov 22 '18 at 15:05
add a comment |
1 Answer
1
active
oldest
votes
Thanks to @MaximEgorushkin for noting that I should look inside usr_input().
The error is in usr_input() and not in user_tui().
It has a very long switch with many switches inside, and in one of them I forgot the default: entry.
So lesson: look recursively inside functions to see if they are really initialized.
I think gcc should let us know that!
answered Nov 22 '18 at 14:01
Cacahuete FritoCacahuete Frito
349317
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53432203%2fspurious-warning-gcc-wuninitialized%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks to @MaximEgorushkin for noting that I should look inside usr_input().
The error is in usr_input() and not in user_tui().
It has a very long switch with many switches inside, and in one of them I forgot the default: entry.
So lesson: look recursively inside functions to see if they are really initialized.
I think gcc should let us know that!
answered Nov 22 '18 at 14:01
Cacahuete FritoCacahuete Frito
349317
add a comment |
Thanks to @MaximEgorushkin for noting that I should look inside usr_input().
The error is in usr_input() and not in user_tui().
It has a very long switch with many switches inside, and in one of them I forgot the default: entry.
So lesson: look recursively inside functions to see if they are really initialized.
I think gcc should let us know that!
answered Nov 22 '18 at 14:01
Cacahuete FritoCacahuete Frito
349317
add a comment |
Thanks to @MaximEgorushkin for noting that I should look inside usr_input().
The error is in usr_input() and not in user_tui().
It has a very long switch with many switches inside, and in one of them I forgot the default: entry.
So lesson: look recursively inside functions to see if they are really initialized.
I think gcc should let us know that!
answered Nov 22 '18 at 14:01
Cacahuete FritoCacahuete Frito
349317
Thanks to @MaximEgorushkin for noting that I should look inside usr_input().
The error is in usr_input() and not in user_tui().
It has a very long switch with many switches inside, and in one of them I forgot the default: entry.
So lesson: look recursively inside functions to see if they are really initialized.
I think gcc should let us know that!
answered Nov 22 '18 at 14:01
Cacahuete FritoCacahuete Frito
349317
answered Nov 22 '18 at 14:01
Cacahuete FritoCacahuete Frito
349317
answered Nov 22 '18 at 14:01
Cacahuete FritoCacahuete Frito
349317
answered Nov 22 '18 at 14:01
Cacahuete FritoCacahuete Frito
349317
349317
add a comment |
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53432203%2fspurious-warning-gcc-wuninitialized%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
Show the definition of
usr_inputfunction.– Maxim Egorushkin
Nov 22 '18 at 13:46
Really no conditions, also not in usr_input() itself, maybe by using exceptions which running out of the initializing scope? BTW "gcc 8.2 has been released as stable, so it shouldn't have many bugs." is definitly wrong if you look in the bugtracker. There are "some" open bugs, some are open and unfixed for multiple major versions... But you are right, such an easy to see bug I also would not expect.
– Klaus
Nov 22 '18 at 13:49
1
@Maxim You were right, I had to look inside
usr_input(). But I would say that the variable returned byusr_input()is the one that is being used uninitialized, and notactionitself.actionis really being initialized, although with garbage in the case of an uninitialized return ofusr_input(). I would call that a bug in gcc.– Cacahuete Frito
Nov 22 '18 at 14:32
Please don't change the topic. Instead use the
Acceptbutton at the answer that solved your problem. That's the way how StackOverflow works. It's also okay toacceptthe own answer.– harper
Nov 22 '18 at 15:04
Ok. However, as the answer is mine, I can't until 2 days from now.
– Cacahuete Frito
Nov 22 '18 at 15:05