How can I create a constant array with some calculation
0
I have code
int a[5];
for (int i = 0; i < 5; i++){
a[i] = i * i;
}
Is there a way to make this array constant so that other code can use it but not change it.
c++
edited Nov 13 '18 at 16:44
Broman
6,257112241
asked Nov 12 '18 at 22:34
Lu J.Lu J.
114
|
show 4 more comments
0
I have code
int a[5];
for (int i = 0; i < 5; i++){
a[i] = i * i;
}
Is there a way to make this array constant so that other code can use it but not change it.
c++
edited Nov 13 '18 at 16:44
Broman
6,257112241
asked Nov 12 '18 at 22:34
Lu J.Lu J.
114
constexpr std::array- go from there. Don't use C-style arrays. Also; The language hasconst_castso you can never be 100% certain that noone will do nasty things to yourconstthing (unfortunately).
– Jesper Juhl
Nov 12 '18 at 22:37
7
C or C++? Pick one.
– Broman
Nov 12 '18 at 22:37
Detail: C does not have constant arrays. C does hasconstarrays. Accepting to change an element of aconstarray is undefined behavior. It might work, might, not, might kill the code.
– chux
Nov 12 '18 at 22:49
2
Is this literally the code you have? What's wrong withconst int a[5] = { 0, 1, 2, 4, 8 };?
– paddy
Nov 12 '18 at 23:03
Lu J., I think @paddy meantconst int a[5] = { 0, 1, 4, 9, 16 };.
– chux
Nov 13 '18 at 2:19
|
show 4 more comments
0
0
0
I have code
int a[5];
for (int i = 0; i < 5; i++){
a[i] = i * i;
}
Is there a way to make this array constant so that other code can use it but not change it.
c++
edited Nov 13 '18 at 16:44
Broman
6,257112241
asked Nov 12 '18 at 22:34
Lu J.Lu J.
114
I have code
int a[5];
for (int i = 0; i < 5; i++){
a[i] = i * i;
}
Is there a way to make this array constant so that other code can use it but not change it.
c++
c++
edited Nov 13 '18 at 16:44
Broman
6,257112241
asked Nov 12 '18 at 22:34
Lu J.Lu J.
114
edited Nov 13 '18 at 16:44
Broman
6,257112241
asked Nov 12 '18 at 22:34
Lu J.Lu J.
114
edited Nov 13 '18 at 16:44
Broman
6,257112241
edited Nov 13 '18 at 16:44
Broman
6,257112241
edited Nov 13 '18 at 16:44
Broman
6,257112241
6,257112241
asked Nov 12 '18 at 22:34
Lu J.Lu J.
114
asked Nov 12 '18 at 22:34
Lu J.Lu J.
114
asked Nov 12 '18 at 22:34
Lu J.Lu J.
114
114
constexpr std::array- go from there. Don't use C-style arrays. Also; The language hasconst_castso you can never be 100% certain that noone will do nasty things to yourconstthing (unfortunately).
– Jesper Juhl
Nov 12 '18 at 22:37
7
C or C++? Pick one.
– Broman
Nov 12 '18 at 22:37
Detail: C does not have constant arrays. C does hasconstarrays. Accepting to change an element of aconstarray is undefined behavior. It might work, might, not, might kill the code.
– chux
Nov 12 '18 at 22:49
2
Is this literally the code you have? What's wrong withconst int a[5] = { 0, 1, 2, 4, 8 };?
– paddy
Nov 12 '18 at 23:03
Lu J., I think @paddy meantconst int a[5] = { 0, 1, 4, 9, 16 };.
– chux
Nov 13 '18 at 2:19
|
show 4 more comments
constexpr std::array- go from there. Don't use C-style arrays. Also; The language hasconst_castso you can never be 100% certain that noone will do nasty things to yourconstthing (unfortunately).
– Jesper Juhl
Nov 12 '18 at 22:37
7
C or C++? Pick one.
– Broman
Nov 12 '18 at 22:37
Detail: C does not have constant arrays. C does hasconstarrays. Accepting to change an element of aconstarray is undefined behavior. It might work, might, not, might kill the code.
– chux
Nov 12 '18 at 22:49
2
Is this literally the code you have? What's wrong withconst int a[5] = { 0, 1, 2, 4, 8 };?
– paddy
Nov 12 '18 at 23:03
Lu J., I think @paddy meantconst int a[5] = { 0, 1, 4, 9, 16 };.
– chux
Nov 13 '18 at 2:19
constexpr std::array - go from there. Don't use C-style arrays. Also; The language has const_cast so you can never be 100% certain that noone will do nasty things to your const thing (unfortunately).– Jesper Juhl
Nov 12 '18 at 22:37
constexpr std::array - go from there. Don't use C-style arrays. Also; The language has const_cast so you can never be 100% certain that noone will do nasty things to your const thing (unfortunately).– Jesper Juhl
Nov 12 '18 at 22:37
7
7
C or C++? Pick one.
– Broman
Nov 12 '18 at 22:37
C or C++? Pick one.
– Broman
Nov 12 '18 at 22:37
Detail: C does not have constant arrays. C does has
const arrays. Accepting to change an element of a const array is undefined behavior. It might work, might, not, might kill the code.– chux
Nov 12 '18 at 22:49
Detail: C does not have constant arrays. C does has
const arrays. Accepting to change an element of a const array is undefined behavior. It might work, might, not, might kill the code.– chux
Nov 12 '18 at 22:49
2
2
Is this literally the code you have? What's wrong with
const int a[5] = { 0, 1, 2, 4, 8 }; ?– paddy
Nov 12 '18 at 23:03
Is this literally the code you have? What's wrong with
const int a[5] = { 0, 1, 2, 4, 8 }; ?– paddy
Nov 12 '18 at 23:03
Lu J., I think @paddy meant
const int a[5] = { 0, 1, 4, 9, 16 };.– chux
Nov 13 '18 at 2:19
Lu J., I think @paddy meant
const int a[5] = { 0, 1, 4, 9, 16 };.– chux
Nov 13 '18 at 2:19
|
show 4 more comments
2 Answers
2
active
oldest
votes
1
Easiest way is to use constexpr function and the std::array:
constexpr std::array<int, 5> make_array() {
std::array<int, 5> a{};
for (int i = 0; i < 5; i++){
a[i] = i * i;
}
return a;
}
//...
const std::array<int, 5> a = make_array();
NB. As noted by @M.M, this code is only valid for C++17, as pre-C++17 the operator on array wasn't constexpr.
answered Nov 12 '18 at 22:40
SergeyASergeyA
42.1k53784
Constexpr isn't really needed here. The jist of solution is the lifting of the initialization into a function.
– NathanOliver
Nov 12 '18 at 22:47
note: this code is undefined behaviour NDR prior to C++17, whenoperatorwas made constexpr
– M.M
Nov 12 '18 at 23:00
1
@NathanOliver it does make a difference. For example, ifais a global variable, on gcc absence of constexpr generates run-time initialization, and with constexpr the aray is statically initialized.
– SergeyA
Nov 12 '18 at 23:02
@M.M no diagnostic on godbolt for either of compilers: gcc.godbolt.org/z/I1MLLS
– SergeyA
Nov 12 '18 at 23:04
1
@M.M ah! Different dialect. I always use C++17 nowadays.
– SergeyA
Nov 12 '18 at 23:04
add a comment |
0
Using modern C++ (array and an Immediately Invoked Lambda (IIL)), this can be achieved:
const auto a = ()
{
std::array< int, 5 > x;
for (int i = 0; i < 5; i++)
{
x[i] = i * i;
}
return x;
}(); // IIL
Using a lambda has benefits over a function call, that you can capture the local variables into the lambda and use them without the values passing through parameters ([&] or [=] instead of ).
Just like a function the lambda can easily be inlined into your code, so there can be no overhead.
answered Nov 12 '18 at 23:13
Robert AndrzejukRobert Andrzejuk
2,71521324
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
});
}
});
draft saved
draft discarded
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%2f53271076%2fhow-can-i-create-a-constant-array-with-some-calculation%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
1
Easiest way is to use constexpr function and the std::array:
constexpr std::array<int, 5> make_array() {
std::array<int, 5> a{};
for (int i = 0; i < 5; i++){
a[i] = i * i;
}
return a;
}
//...
const std::array<int, 5> a = make_array();
NB. As noted by @M.M, this code is only valid for C++17, as pre-C++17 the operator on array wasn't constexpr.
answered Nov 12 '18 at 22:40
SergeyASergeyA
42.1k53784
Constexpr isn't really needed here. The jist of solution is the lifting of the initialization into a function.
– NathanOliver
Nov 12 '18 at 22:47
note: this code is undefined behaviour NDR prior to C++17, whenoperatorwas made constexpr
– M.M
Nov 12 '18 at 23:00
1
@NathanOliver it does make a difference. For example, ifais a global variable, on gcc absence of constexpr generates run-time initialization, and with constexpr the aray is statically initialized.
– SergeyA
Nov 12 '18 at 23:02
@M.M no diagnostic on godbolt for either of compilers: gcc.godbolt.org/z/I1MLLS
– SergeyA
Nov 12 '18 at 23:04
1
@M.M ah! Different dialect. I always use C++17 nowadays.
– SergeyA
Nov 12 '18 at 23:04
add a comment |
1
Easiest way is to use constexpr function and the std::array:
constexpr std::array<int, 5> make_array() {
std::array<int, 5> a{};
for (int i = 0; i < 5; i++){
a[i] = i * i;
}
return a;
}
//...
const std::array<int, 5> a = make_array();
NB. As noted by @M.M, this code is only valid for C++17, as pre-C++17 the operator on array wasn't constexpr.
answered Nov 12 '18 at 22:40
SergeyASergeyA
42.1k53784
Constexpr isn't really needed here. The jist of solution is the lifting of the initialization into a function.
– NathanOliver
Nov 12 '18 at 22:47
note: this code is undefined behaviour NDR prior to C++17, whenoperatorwas made constexpr
– M.M
Nov 12 '18 at 23:00
1
@NathanOliver it does make a difference. For example, ifais a global variable, on gcc absence of constexpr generates run-time initialization, and with constexpr the aray is statically initialized.
– SergeyA
Nov 12 '18 at 23:02
@M.M no diagnostic on godbolt for either of compilers: gcc.godbolt.org/z/I1MLLS
– SergeyA
Nov 12 '18 at 23:04
1
@M.M ah! Different dialect. I always use C++17 nowadays.
– SergeyA
Nov 12 '18 at 23:04
add a comment |
1
1
1
Easiest way is to use constexpr function and the std::array:
constexpr std::array<int, 5> make_array() {
std::array<int, 5> a{};
for (int i = 0; i < 5; i++){
a[i] = i * i;
}
return a;
}
//...
const std::array<int, 5> a = make_array();
NB. As noted by @M.M, this code is only valid for C++17, as pre-C++17 the operator on array wasn't constexpr.
answered Nov 12 '18 at 22:40
SergeyASergeyA
42.1k53784
Easiest way is to use constexpr function and the std::array:
constexpr std::array<int, 5> make_array() {
std::array<int, 5> a{};
for (int i = 0; i < 5; i++){
a[i] = i * i;
}
return a;
}
//...
const std::array<int, 5> a = make_array();
NB. As noted by @M.M, this code is only valid for C++17, as pre-C++17 the operator on array wasn't constexpr.
answered Nov 12 '18 at 22:40
SergeyASergeyA
42.1k53784
edited Nov 12 '18 at 23:05
answered Nov 12 '18 at 22:40
SergeyASergeyA
42.1k53784
answered Nov 12 '18 at 22:40
SergeyASergeyA
42.1k53784
answered Nov 12 '18 at 22:40
SergeyASergeyA
42.1k53784
42.1k53784
Constexpr isn't really needed here. The jist of solution is the lifting of the initialization into a function.
– NathanOliver
Nov 12 '18 at 22:47
note: this code is undefined behaviour NDR prior to C++17, whenoperatorwas made constexpr
– M.M
Nov 12 '18 at 23:00
1
@NathanOliver it does make a difference. For example, ifais a global variable, on gcc absence of constexpr generates run-time initialization, and with constexpr the aray is statically initialized.
– SergeyA
Nov 12 '18 at 23:02
@M.M no diagnostic on godbolt for either of compilers: gcc.godbolt.org/z/I1MLLS
– SergeyA
Nov 12 '18 at 23:04
1
@M.M ah! Different dialect. I always use C++17 nowadays.
– SergeyA
Nov 12 '18 at 23:04
add a comment |
Constexpr isn't really needed here. The jist of solution is the lifting of the initialization into a function.
– NathanOliver
Nov 12 '18 at 22:47
note: this code is undefined behaviour NDR prior to C++17, whenoperatorwas made constexpr
– M.M
Nov 12 '18 at 23:00
1
@NathanOliver it does make a difference. For example, ifais a global variable, on gcc absence of constexpr generates run-time initialization, and with constexpr the aray is statically initialized.
– SergeyA
Nov 12 '18 at 23:02
@M.M no diagnostic on godbolt for either of compilers: gcc.godbolt.org/z/I1MLLS
– SergeyA
Nov 12 '18 at 23:04
1
@M.M ah! Different dialect. I always use C++17 nowadays.
– SergeyA
Nov 12 '18 at 23:04
Constexpr isn't really needed here. The jist of solution is the lifting of the initialization into a function.
– NathanOliver
Nov 12 '18 at 22:47
Constexpr isn't really needed here. The jist of solution is the lifting of the initialization into a function.
– NathanOliver
Nov 12 '18 at 22:47
note: this code is undefined behaviour NDR prior to C++17, when
operator was made constexpr– M.M
Nov 12 '18 at 23:00
note: this code is undefined behaviour NDR prior to C++17, when
operator was made constexpr– M.M
Nov 12 '18 at 23:00
1
1
@NathanOliver it does make a difference. For example, if
a is a global variable, on gcc absence of constexpr generates run-time initialization, and with constexpr the aray is statically initialized.– SergeyA
Nov 12 '18 at 23:02
@NathanOliver it does make a difference. For example, if
a is a global variable, on gcc absence of constexpr generates run-time initialization, and with constexpr the aray is statically initialized.– SergeyA
Nov 12 '18 at 23:02
@M.M no diagnostic on godbolt for either of compilers: gcc.godbolt.org/z/I1MLLS
– SergeyA
Nov 12 '18 at 23:04
@M.M no diagnostic on godbolt for either of compilers: gcc.godbolt.org/z/I1MLLS
– SergeyA
Nov 12 '18 at 23:04
1
1
@M.M ah! Different dialect. I always use C++17 nowadays.
– SergeyA
Nov 12 '18 at 23:04
@M.M ah! Different dialect. I always use C++17 nowadays.
– SergeyA
Nov 12 '18 at 23:04
add a comment |
0
Using modern C++ (array and an Immediately Invoked Lambda (IIL)), this can be achieved:
const auto a = ()
{
std::array< int, 5 > x;
for (int i = 0; i < 5; i++)
{
x[i] = i * i;
}
return x;
}(); // IIL
Using a lambda has benefits over a function call, that you can capture the local variables into the lambda and use them without the values passing through parameters ([&] or [=] instead of ).
Just like a function the lambda can easily be inlined into your code, so there can be no overhead.
answered Nov 12 '18 at 23:13
Robert AndrzejukRobert Andrzejuk
2,71521324
add a comment |
0
Using modern C++ (array and an Immediately Invoked Lambda (IIL)), this can be achieved:
const auto a = ()
{
std::array< int, 5 > x;
for (int i = 0; i < 5; i++)
{
x[i] = i * i;
}
return x;
}(); // IIL
Using a lambda has benefits over a function call, that you can capture the local variables into the lambda and use them without the values passing through parameters ([&] or [=] instead of ).
Just like a function the lambda can easily be inlined into your code, so there can be no overhead.
answered Nov 12 '18 at 23:13
Robert AndrzejukRobert Andrzejuk
2,71521324
add a comment |
0
0
0
Using modern C++ (array and an Immediately Invoked Lambda (IIL)), this can be achieved:
const auto a = ()
{
std::array< int, 5 > x;
for (int i = 0; i < 5; i++)
{
x[i] = i * i;
}
return x;
}(); // IIL
Using a lambda has benefits over a function call, that you can capture the local variables into the lambda and use them without the values passing through parameters ([&] or [=] instead of ).
Just like a function the lambda can easily be inlined into your code, so there can be no overhead.
answered Nov 12 '18 at 23:13
Robert AndrzejukRobert Andrzejuk
2,71521324
Using modern C++ (array and an Immediately Invoked Lambda (IIL)), this can be achieved:
const auto a = ()
{
std::array< int, 5 > x;
for (int i = 0; i < 5; i++)
{
x[i] = i * i;
}
return x;
}(); // IIL
Using a lambda has benefits over a function call, that you can capture the local variables into the lambda and use them without the values passing through parameters ([&] or [=] instead of ).
Just like a function the lambda can easily be inlined into your code, so there can be no overhead.
answered Nov 12 '18 at 23:13
Robert AndrzejukRobert Andrzejuk
2,71521324
edited Nov 14 '18 at 12:42
answered Nov 12 '18 at 23:13
Robert AndrzejukRobert Andrzejuk
2,71521324
answered Nov 12 '18 at 23:13
Robert AndrzejukRobert Andrzejuk
2,71521324
answered Nov 12 '18 at 23:13
Robert AndrzejukRobert Andrzejuk
2,71521324
2,71521324
add a comment |
add a comment |
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f53271076%2fhow-can-i-create-a-constant-array-with-some-calculation%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
constexpr std::array- go from there. Don't use C-style arrays. Also; The language hasconst_castso you can never be 100% certain that noone will do nasty things to yourconstthing (unfortunately).– Jesper Juhl
Nov 12 '18 at 22:37
7
C or C++? Pick one.
– Broman
Nov 12 '18 at 22:37
Detail: C does not have constant arrays. C does has
constarrays. Accepting to change an element of aconstarray is undefined behavior. It might work, might, not, might kill the code.– chux
Nov 12 '18 at 22:49
2
Is this literally the code you have? What's wrong with
const int a[5] = { 0, 1, 2, 4, 8 };?– paddy
Nov 12 '18 at 23:03
Lu J., I think @paddy meant
const int a[5] = { 0, 1, 4, 9, 16 };.– chux
Nov 13 '18 at 2:19