In C++14 is it valid to use a double in the dimension of a new expression?
In C++14 given the following code:
void foo() {
double d = 5.0;
auto p1 = new int[d];
}
clang compiles this without diagnostic while gcc on the other hand produces the following diagnostic (see it live in godbolt):
error: expression in new-declarator must have integral or enumeration type
7 | auto p1 = new int[d];
| ^
I specifically labeled this C++14 because in C++11 mode clang treats this as ill-formed and produces the following diagnostic (see it live in godbolt):
error: array size expression must have integral or unscoped enumeration type, not 'double'
auto p1 = new int[d];
^ ~
Is clang correct? If so what changed in C++14 to allow this?
c++ c++14 language-lawyer new-expression
asked Dec 12 '18 at 14:25
Shafik YaghmourShafik Yaghmour
127k23329546
|
show 4 more comments
In C++14 given the following code:
void foo() {
double d = 5.0;
auto p1 = new int[d];
}
clang compiles this without diagnostic while gcc on the other hand produces the following diagnostic (see it live in godbolt):
error: expression in new-declarator must have integral or enumeration type
7 | auto p1 = new int[d];
| ^
I specifically labeled this C++14 because in C++11 mode clang treats this as ill-formed and produces the following diagnostic (see it live in godbolt):
error: array size expression must have integral or unscoped enumeration type, not 'double'
auto p1 = new int[d];
^ ~
Is clang correct? If so what changed in C++14 to allow this?
c++ c++14 language-lawyer new-expression
asked Dec 12 '18 at 14:25
Shafik YaghmourShafik Yaghmour
127k23329546
2
Out of curiosity, a double allows for fractional quantities, so how would you allocate 0.75 of an integer, such asint * p_array = new int [0.75];? Or take something like 0.33333333, which is kind of difficult to allocate.
– Thomas Matthews
Dec 12 '18 at 15:13
5
@ThomasMatthews no this would end up being a float to integral conversion and would truncate the float
– Shafik Yaghmour
Dec 12 '18 at 15:59
1
@tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)
– Bob__
Dec 13 '18 at 9:50
1
@tomerzeitune Whyint? Why notunsignedorlong?
– curiousguy
Dec 13 '18 at 22:35
1
@KeithThompson I think the commenters are joking around
– Shafik Yaghmour
Dec 14 '18 at 18:11
|
show 4 more comments
In C++14 given the following code:
void foo() {
double d = 5.0;
auto p1 = new int[d];
}
clang compiles this without diagnostic while gcc on the other hand produces the following diagnostic (see it live in godbolt):
error: expression in new-declarator must have integral or enumeration type
7 | auto p1 = new int[d];
| ^
I specifically labeled this C++14 because in C++11 mode clang treats this as ill-formed and produces the following diagnostic (see it live in godbolt):
error: array size expression must have integral or unscoped enumeration type, not 'double'
auto p1 = new int[d];
^ ~
Is clang correct? If so what changed in C++14 to allow this?
c++ c++14 language-lawyer new-expression
asked Dec 12 '18 at 14:25
Shafik YaghmourShafik Yaghmour
127k23329546
In C++14 given the following code:
void foo() {
double d = 5.0;
auto p1 = new int[d];
}
clang compiles this without diagnostic while gcc on the other hand produces the following diagnostic (see it live in godbolt):
error: expression in new-declarator must have integral or enumeration type
7 | auto p1 = new int[d];
| ^
I specifically labeled this C++14 because in C++11 mode clang treats this as ill-formed and produces the following diagnostic (see it live in godbolt):
error: array size expression must have integral or unscoped enumeration type, not 'double'
auto p1 = new int[d];
^ ~
Is clang correct? If so what changed in C++14 to allow this?
c++ c++14 language-lawyer new-expression
c++ c++14 language-lawyer new-expression
asked Dec 12 '18 at 14:25
Shafik YaghmourShafik Yaghmour
127k23329546
asked Dec 12 '18 at 14:25
Shafik YaghmourShafik Yaghmour
127k23329546
edited Dec 13 '18 at 21:10
Shafik Yaghmour
asked Dec 12 '18 at 14:25
Shafik YaghmourShafik Yaghmour
127k23329546
asked Dec 12 '18 at 14:25
Shafik YaghmourShafik Yaghmour
127k23329546
asked Dec 12 '18 at 14:25
Shafik YaghmourShafik Yaghmour
127k23329546
127k23329546
2
Out of curiosity, a double allows for fractional quantities, so how would you allocate 0.75 of an integer, such asint * p_array = new int [0.75];? Or take something like 0.33333333, which is kind of difficult to allocate.
– Thomas Matthews
Dec 12 '18 at 15:13
5
@ThomasMatthews no this would end up being a float to integral conversion and would truncate the float
– Shafik Yaghmour
Dec 12 '18 at 15:59
1
@tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)
– Bob__
Dec 13 '18 at 9:50
1
@tomerzeitune Whyint? Why notunsignedorlong?
– curiousguy
Dec 13 '18 at 22:35
1
@KeithThompson I think the commenters are joking around
– Shafik Yaghmour
Dec 14 '18 at 18:11
|
show 4 more comments
2
Out of curiosity, a double allows for fractional quantities, so how would you allocate 0.75 of an integer, such asint * p_array = new int [0.75];? Or take something like 0.33333333, which is kind of difficult to allocate.
– Thomas Matthews
Dec 12 '18 at 15:13
5
@ThomasMatthews no this would end up being a float to integral conversion and would truncate the float
– Shafik Yaghmour
Dec 12 '18 at 15:59
1
@tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)
– Bob__
Dec 13 '18 at 9:50
1
@tomerzeitune Whyint? Why notunsignedorlong?
– curiousguy
Dec 13 '18 at 22:35
1
@KeithThompson I think the commenters are joking around
– Shafik Yaghmour
Dec 14 '18 at 18:11
2
2
Out of curiosity, a double allows for fractional quantities, so how would you allocate 0.75 of an integer, such as
int * p_array = new int [0.75];? Or take something like 0.33333333, which is kind of difficult to allocate.– Thomas Matthews
Dec 12 '18 at 15:13
Out of curiosity, a double allows for fractional quantities, so how would you allocate 0.75 of an integer, such as
int * p_array = new int [0.75];? Or take something like 0.33333333, which is kind of difficult to allocate.– Thomas Matthews
Dec 12 '18 at 15:13
5
5
@ThomasMatthews no this would end up being a float to integral conversion and would truncate the float
– Shafik Yaghmour
Dec 12 '18 at 15:59
@ThomasMatthews no this would end up being a float to integral conversion and would truncate the float
– Shafik Yaghmour
Dec 12 '18 at 15:59
1
1
@tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)
– Bob__
Dec 13 '18 at 9:50
@tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)
– Bob__
Dec 13 '18 at 9:50
1
1
@tomerzeitune Why
int? Why not unsigned or long?– curiousguy
Dec 13 '18 at 22:35
@tomerzeitune Why
int? Why not unsigned or long?– curiousguy
Dec 13 '18 at 22:35
1
1
@KeithThompson I think the commenters are joking around
– Shafik Yaghmour
Dec 14 '18 at 18:11
@KeithThompson I think the commenters are joking around
– Shafik Yaghmour
Dec 14 '18 at 18:11
|
show 4 more comments
2 Answers
2
active
oldest
votes
Clang is correct, the key wording in [expr.new]p6 changes from the following in the C++11 draft:
Every constant-expression in a noptr-new-declarator shall be an integral constant expression ([expr.const]) and evaluate to a strictly positive value. The expression in a noptr-new-declarator shall be of integral type, unscoped enumeration type, or a class type for which a single non-explicit conversion function to integral or unscoped enumeration type exists ([class.conv]). If the expression is of class type, the expression is converted by calling that conversion function, and the result of the conversion is used in place of the original expression. …
to this in the C++14 draft:
Every constant-expression in a noptr-new-declarator shall be a converted constant expression ([expr.const]) of type
std::size_tand shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted tostd::size_t. …
In C++14 the requirement for the expression in a noptr-new-declarator was weakened to not require an integral, unscoped enumeration or a class with a
single non-explicit conversion function to one of those types but just allow implicit conversions to size_t.
The change in wording came from the proposal A Proposal to Tweak Certain C++ Contextual Conversions, v3.
answered Dec 12 '18 at 14:25
Shafik YaghmourShafik Yaghmour
127k23329546
19
I am dubious about the usefulness of allowingdoubleto be used as the size of an array... it seems more likely to let bugs pass silently than anything else :(
– Matthieu M.
Dec 12 '18 at 15:48
12
@MatthieuM. I agree, I believe it is a defect and that the intent was really to saycontextually implicitly converted. I am filing a defect report on this hopefully today but who knows maybe I am wrong :-(
– Shafik Yaghmour
Dec 12 '18 at 15:57
2
@MatthieuM. FYI I filed a defect report, processing takes a while so I don't expect to have an update anytime soon though.
– Shafik Yaghmour
Dec 14 '18 at 17:56
add a comment |
From c++14 to c++17 (for the ones that wonder like me), the phrasing remains practically the same (unlike from C++11 to C++14 as @ShafikYaghmour answered), as stated in this C++17 draft:
Every constant-expression in a noptr-new-declarator shall be a converted constant expression of type
std::size_tand shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted tostd::size_t. [..]
with only this part ([expr.const]) missing from the C++17 draft.
edited Jan 10 at 8:42
Bakudan
14k84266
answered Dec 14 '18 at 14:34
gsamarasgsamaras
52.3k25107193
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%2f53745158%2fin-c14-is-it-valid-to-use-a-double-in-the-dimension-of-a-new-expression%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Clang is correct, the key wording in [expr.new]p6 changes from the following in the C++11 draft:
Every constant-expression in a noptr-new-declarator shall be an integral constant expression ([expr.const]) and evaluate to a strictly positive value. The expression in a noptr-new-declarator shall be of integral type, unscoped enumeration type, or a class type for which a single non-explicit conversion function to integral or unscoped enumeration type exists ([class.conv]). If the expression is of class type, the expression is converted by calling that conversion function, and the result of the conversion is used in place of the original expression. …
to this in the C++14 draft:
Every constant-expression in a noptr-new-declarator shall be a converted constant expression ([expr.const]) of type
std::size_tand shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted tostd::size_t. …
In C++14 the requirement for the expression in a noptr-new-declarator was weakened to not require an integral, unscoped enumeration or a class with a
single non-explicit conversion function to one of those types but just allow implicit conversions to size_t.
The change in wording came from the proposal A Proposal to Tweak Certain C++ Contextual Conversions, v3.
answered Dec 12 '18 at 14:25
Shafik YaghmourShafik Yaghmour
127k23329546
19
I am dubious about the usefulness of allowingdoubleto be used as the size of an array... it seems more likely to let bugs pass silently than anything else :(
– Matthieu M.
Dec 12 '18 at 15:48
12
@MatthieuM. I agree, I believe it is a defect and that the intent was really to saycontextually implicitly converted. I am filing a defect report on this hopefully today but who knows maybe I am wrong :-(
– Shafik Yaghmour
Dec 12 '18 at 15:57
2
@MatthieuM. FYI I filed a defect report, processing takes a while so I don't expect to have an update anytime soon though.
– Shafik Yaghmour
Dec 14 '18 at 17:56
add a comment |
Clang is correct, the key wording in [expr.new]p6 changes from the following in the C++11 draft:
Every constant-expression in a noptr-new-declarator shall be an integral constant expression ([expr.const]) and evaluate to a strictly positive value. The expression in a noptr-new-declarator shall be of integral type, unscoped enumeration type, or a class type for which a single non-explicit conversion function to integral or unscoped enumeration type exists ([class.conv]). If the expression is of class type, the expression is converted by calling that conversion function, and the result of the conversion is used in place of the original expression. …
to this in the C++14 draft:
Every constant-expression in a noptr-new-declarator shall be a converted constant expression ([expr.const]) of type
std::size_tand shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted tostd::size_t. …
In C++14 the requirement for the expression in a noptr-new-declarator was weakened to not require an integral, unscoped enumeration or a class with a
single non-explicit conversion function to one of those types but just allow implicit conversions to size_t.
The change in wording came from the proposal A Proposal to Tweak Certain C++ Contextual Conversions, v3.
answered Dec 12 '18 at 14:25
Shafik YaghmourShafik Yaghmour
127k23329546
19
I am dubious about the usefulness of allowingdoubleto be used as the size of an array... it seems more likely to let bugs pass silently than anything else :(
– Matthieu M.
Dec 12 '18 at 15:48
12
@MatthieuM. I agree, I believe it is a defect and that the intent was really to saycontextually implicitly converted. I am filing a defect report on this hopefully today but who knows maybe I am wrong :-(
– Shafik Yaghmour
Dec 12 '18 at 15:57
2
@MatthieuM. FYI I filed a defect report, processing takes a while so I don't expect to have an update anytime soon though.
– Shafik Yaghmour
Dec 14 '18 at 17:56
add a comment |
Clang is correct, the key wording in [expr.new]p6 changes from the following in the C++11 draft:
Every constant-expression in a noptr-new-declarator shall be an integral constant expression ([expr.const]) and evaluate to a strictly positive value. The expression in a noptr-new-declarator shall be of integral type, unscoped enumeration type, or a class type for which a single non-explicit conversion function to integral or unscoped enumeration type exists ([class.conv]). If the expression is of class type, the expression is converted by calling that conversion function, and the result of the conversion is used in place of the original expression. …
to this in the C++14 draft:
Every constant-expression in a noptr-new-declarator shall be a converted constant expression ([expr.const]) of type
std::size_tand shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted tostd::size_t. …
In C++14 the requirement for the expression in a noptr-new-declarator was weakened to not require an integral, unscoped enumeration or a class with a
single non-explicit conversion function to one of those types but just allow implicit conversions to size_t.
The change in wording came from the proposal A Proposal to Tweak Certain C++ Contextual Conversions, v3.
answered Dec 12 '18 at 14:25
Shafik YaghmourShafik Yaghmour
127k23329546
Clang is correct, the key wording in [expr.new]p6 changes from the following in the C++11 draft:
Every constant-expression in a noptr-new-declarator shall be an integral constant expression ([expr.const]) and evaluate to a strictly positive value. The expression in a noptr-new-declarator shall be of integral type, unscoped enumeration type, or a class type for which a single non-explicit conversion function to integral or unscoped enumeration type exists ([class.conv]). If the expression is of class type, the expression is converted by calling that conversion function, and the result of the conversion is used in place of the original expression. …
to this in the C++14 draft:
Every constant-expression in a noptr-new-declarator shall be a converted constant expression ([expr.const]) of type
std::size_tand shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted tostd::size_t. …
In C++14 the requirement for the expression in a noptr-new-declarator was weakened to not require an integral, unscoped enumeration or a class with a
single non-explicit conversion function to one of those types but just allow implicit conversions to size_t.
The change in wording came from the proposal A Proposal to Tweak Certain C++ Contextual Conversions, v3.
answered Dec 12 '18 at 14:25
Shafik YaghmourShafik Yaghmour
127k23329546
edited Dec 15 '18 at 15:00
answered Dec 12 '18 at 14:25
Shafik YaghmourShafik Yaghmour
127k23329546
answered Dec 12 '18 at 14:25
Shafik YaghmourShafik Yaghmour
127k23329546
answered Dec 12 '18 at 14:25
Shafik YaghmourShafik Yaghmour
127k23329546
127k23329546
19
I am dubious about the usefulness of allowingdoubleto be used as the size of an array... it seems more likely to let bugs pass silently than anything else :(
– Matthieu M.
Dec 12 '18 at 15:48
12
@MatthieuM. I agree, I believe it is a defect and that the intent was really to saycontextually implicitly converted. I am filing a defect report on this hopefully today but who knows maybe I am wrong :-(
– Shafik Yaghmour
Dec 12 '18 at 15:57
2
@MatthieuM. FYI I filed a defect report, processing takes a while so I don't expect to have an update anytime soon though.
– Shafik Yaghmour
Dec 14 '18 at 17:56
add a comment |
19
I am dubious about the usefulness of allowingdoubleto be used as the size of an array... it seems more likely to let bugs pass silently than anything else :(
– Matthieu M.
Dec 12 '18 at 15:48
12
@MatthieuM. I agree, I believe it is a defect and that the intent was really to saycontextually implicitly converted. I am filing a defect report on this hopefully today but who knows maybe I am wrong :-(
– Shafik Yaghmour
Dec 12 '18 at 15:57
2
@MatthieuM. FYI I filed a defect report, processing takes a while so I don't expect to have an update anytime soon though.
– Shafik Yaghmour
Dec 14 '18 at 17:56
19
19
I am dubious about the usefulness of allowing
double to be used as the size of an array... it seems more likely to let bugs pass silently than anything else :(– Matthieu M.
Dec 12 '18 at 15:48
I am dubious about the usefulness of allowing
double to be used as the size of an array... it seems more likely to let bugs pass silently than anything else :(– Matthieu M.
Dec 12 '18 at 15:48
12
12
@MatthieuM. I agree, I believe it is a defect and that the intent was really to say
contextually implicitly converted. I am filing a defect report on this hopefully today but who knows maybe I am wrong :-(– Shafik Yaghmour
Dec 12 '18 at 15:57
@MatthieuM. I agree, I believe it is a defect and that the intent was really to say
contextually implicitly converted. I am filing a defect report on this hopefully today but who knows maybe I am wrong :-(– Shafik Yaghmour
Dec 12 '18 at 15:57
2
2
@MatthieuM. FYI I filed a defect report, processing takes a while so I don't expect to have an update anytime soon though.
– Shafik Yaghmour
Dec 14 '18 at 17:56
@MatthieuM. FYI I filed a defect report, processing takes a while so I don't expect to have an update anytime soon though.
– Shafik Yaghmour
Dec 14 '18 at 17:56
add a comment |
From c++14 to c++17 (for the ones that wonder like me), the phrasing remains practically the same (unlike from C++11 to C++14 as @ShafikYaghmour answered), as stated in this C++17 draft:
Every constant-expression in a noptr-new-declarator shall be a converted constant expression of type
std::size_tand shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted tostd::size_t. [..]
with only this part ([expr.const]) missing from the C++17 draft.
edited Jan 10 at 8:42
Bakudan
14k84266
answered Dec 14 '18 at 14:34
gsamarasgsamaras
52.3k25107193
add a comment |
From c++14 to c++17 (for the ones that wonder like me), the phrasing remains practically the same (unlike from C++11 to C++14 as @ShafikYaghmour answered), as stated in this C++17 draft:
Every constant-expression in a noptr-new-declarator shall be a converted constant expression of type
std::size_tand shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted tostd::size_t. [..]
with only this part ([expr.const]) missing from the C++17 draft.
edited Jan 10 at 8:42
Bakudan
14k84266
answered Dec 14 '18 at 14:34
gsamarasgsamaras
52.3k25107193
add a comment |
From c++14 to c++17 (for the ones that wonder like me), the phrasing remains practically the same (unlike from C++11 to C++14 as @ShafikYaghmour answered), as stated in this C++17 draft:
Every constant-expression in a noptr-new-declarator shall be a converted constant expression of type
std::size_tand shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted tostd::size_t. [..]
with only this part ([expr.const]) missing from the C++17 draft.
edited Jan 10 at 8:42
Bakudan
14k84266
answered Dec 14 '18 at 14:34
gsamarasgsamaras
52.3k25107193
From c++14 to c++17 (for the ones that wonder like me), the phrasing remains practically the same (unlike from C++11 to C++14 as @ShafikYaghmour answered), as stated in this C++17 draft:
Every constant-expression in a noptr-new-declarator shall be a converted constant expression of type
std::size_tand shall evaluate to a strictly positive value. The expression in a noptr-new-declarator is implicitly converted tostd::size_t. [..]
with only this part ([expr.const]) missing from the C++17 draft.
edited Jan 10 at 8:42
Bakudan
14k84266
answered Dec 14 '18 at 14:34
gsamarasgsamaras
52.3k25107193
edited Jan 10 at 8:42
Bakudan
14k84266
edited Jan 10 at 8:42
Bakudan
14k84266
edited Jan 10 at 8:42
Bakudan
14k84266
14k84266
answered Dec 14 '18 at 14:34
gsamarasgsamaras
52.3k25107193
answered Dec 14 '18 at 14:34
gsamarasgsamaras
52.3k25107193
answered Dec 14 '18 at 14:34
gsamarasgsamaras
52.3k25107193
52.3k25107193
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%2f53745158%2fin-c14-is-it-valid-to-use-a-double-in-the-dimension-of-a-new-expression%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
2
Out of curiosity, a double allows for fractional quantities, so how would you allocate 0.75 of an integer, such as
int * p_array = new int [0.75];? Or take something like 0.33333333, which is kind of difficult to allocate.– Thomas Matthews
Dec 12 '18 at 15:13
5
@ThomasMatthews no this would end up being a float to integral conversion and would truncate the float
– Shafik Yaghmour
Dec 12 '18 at 15:59
1
@tomerzeitune Well, that (0.75 -> 0) is not what you wrote (0.75 -> 16360) ;)
– Bob__
Dec 13 '18 at 9:50
1
@tomerzeitune Why
int? Why notunsignedorlong?– curiousguy
Dec 13 '18 at 22:35
1
@KeithThompson I think the commenters are joking around
– Shafik Yaghmour
Dec 14 '18 at 18:11