Different results when using auto as member function argument
I am getting different results when running these two
I am on GNU/Linux 4.14.67
These both are ran using g++ -std=c++14 with/without -O0 and on c++17 as well.
Why am I? Why are the outputs different?
First version is:
#include <iostream>
#include <algorithm>
using namespace std;
class foo {
public:
foo() { }
foo(const foo& f) { }
foo& operator=(const foo& f) {
cout << "foo operator=n";
val = 888;
// Do something important
return *this;
}
int val;
};
int main() {
foo f1;
foo f2;
f1 = f2;
cout << f1.val << endl;
}
The first output is:
foo operator=
888
Second version (only changing const foo& to const auto& ):
#include <iostream>
#include <algorithm>
using namespace std;
class foo {
public:
foo() { }
foo(const foo& f) { }
foo& operator=(const auto& f) {
cout << "foo operator=n";
val = 888;
// Do something important
return *this;
}
int val;
};
int main() {
foo f1;
foo f2;
f1 = f2;
cout << f1.val << endl;
}
The second output is:
0
c++ c++14 c++17
asked Nov 16 '18 at 4:44
SeoulSeoul
216111
|
show 8 more comments
I am getting different results when running these two
I am on GNU/Linux 4.14.67
These both are ran using g++ -std=c++14 with/without -O0 and on c++17 as well.
Why am I? Why are the outputs different?
First version is:
#include <iostream>
#include <algorithm>
using namespace std;
class foo {
public:
foo() { }
foo(const foo& f) { }
foo& operator=(const foo& f) {
cout << "foo operator=n";
val = 888;
// Do something important
return *this;
}
int val;
};
int main() {
foo f1;
foo f2;
f1 = f2;
cout << f1.val << endl;
}
The first output is:
foo operator=
888
Second version (only changing const foo& to const auto& ):
#include <iostream>
#include <algorithm>
using namespace std;
class foo {
public:
foo() { }
foo(const foo& f) { }
foo& operator=(const auto& f) {
cout << "foo operator=n";
val = 888;
// Do something important
return *this;
}
int val;
};
int main() {
foo f1;
foo f2;
f1 = f2;
cout << f1.val << endl;
}
The second output is:
0
c++ c++14 c++17
asked Nov 16 '18 at 4:44
SeoulSeoul
216111
1
Maybe your forgot "return *this" at the end of op=(...) ?
– Picaud Vincent
Nov 16 '18 at 4:46
I thought so too, but it was a copy and paste error. Good catch
– Seoul
Nov 16 '18 at 4:47
1
Not only "*this" but "return *this;"
– Picaud Vincent
Nov 16 '18 at 4:47
7
"Why am I?" Hard to tell, very philosophical question. Some say "I am thinking, thus I be (am)"
– πάντα ῥεῖ
Nov 16 '18 at 4:48
1
@Swordfish Watchworthy: youtube.com/watch?v=29pPZQ77cmI
– πάντα ῥεῖ
Nov 16 '18 at 4:53
|
show 8 more comments
I am getting different results when running these two
I am on GNU/Linux 4.14.67
These both are ran using g++ -std=c++14 with/without -O0 and on c++17 as well.
Why am I? Why are the outputs different?
First version is:
#include <iostream>
#include <algorithm>
using namespace std;
class foo {
public:
foo() { }
foo(const foo& f) { }
foo& operator=(const foo& f) {
cout << "foo operator=n";
val = 888;
// Do something important
return *this;
}
int val;
};
int main() {
foo f1;
foo f2;
f1 = f2;
cout << f1.val << endl;
}
The first output is:
foo operator=
888
Second version (only changing const foo& to const auto& ):
#include <iostream>
#include <algorithm>
using namespace std;
class foo {
public:
foo() { }
foo(const foo& f) { }
foo& operator=(const auto& f) {
cout << "foo operator=n";
val = 888;
// Do something important
return *this;
}
int val;
};
int main() {
foo f1;
foo f2;
f1 = f2;
cout << f1.val << endl;
}
The second output is:
0
c++ c++14 c++17
asked Nov 16 '18 at 4:44
SeoulSeoul
216111
I am getting different results when running these two
I am on GNU/Linux 4.14.67
These both are ran using g++ -std=c++14 with/without -O0 and on c++17 as well.
Why am I? Why are the outputs different?
First version is:
#include <iostream>
#include <algorithm>
using namespace std;
class foo {
public:
foo() { }
foo(const foo& f) { }
foo& operator=(const foo& f) {
cout << "foo operator=n";
val = 888;
// Do something important
return *this;
}
int val;
};
int main() {
foo f1;
foo f2;
f1 = f2;
cout << f1.val << endl;
}
The first output is:
foo operator=
888
Second version (only changing const foo& to const auto& ):
#include <iostream>
#include <algorithm>
using namespace std;
class foo {
public:
foo() { }
foo(const foo& f) { }
foo& operator=(const auto& f) {
cout << "foo operator=n";
val = 888;
// Do something important
return *this;
}
int val;
};
int main() {
foo f1;
foo f2;
f1 = f2;
cout << f1.val << endl;
}
The second output is:
0
c++ c++14 c++17
c++ c++14 c++17
asked Nov 16 '18 at 4:44
SeoulSeoul
216111
asked Nov 16 '18 at 4:44
SeoulSeoul
216111
edited Nov 16 '18 at 4:48
Seoul
asked Nov 16 '18 at 4:44
SeoulSeoul
216111
asked Nov 16 '18 at 4:44
SeoulSeoul
216111
asked Nov 16 '18 at 4:44
SeoulSeoul
216111
216111
1
Maybe your forgot "return *this" at the end of op=(...) ?
– Picaud Vincent
Nov 16 '18 at 4:46
I thought so too, but it was a copy and paste error. Good catch
– Seoul
Nov 16 '18 at 4:47
1
Not only "*this" but "return *this;"
– Picaud Vincent
Nov 16 '18 at 4:47
7
"Why am I?" Hard to tell, very philosophical question. Some say "I am thinking, thus I be (am)"
– πάντα ῥεῖ
Nov 16 '18 at 4:48
1
@Swordfish Watchworthy: youtube.com/watch?v=29pPZQ77cmI
– πάντα ῥεῖ
Nov 16 '18 at 4:53
|
show 8 more comments
1
Maybe your forgot "return *this" at the end of op=(...) ?
– Picaud Vincent
Nov 16 '18 at 4:46
I thought so too, but it was a copy and paste error. Good catch
– Seoul
Nov 16 '18 at 4:47
1
Not only "*this" but "return *this;"
– Picaud Vincent
Nov 16 '18 at 4:47
7
"Why am I?" Hard to tell, very philosophical question. Some say "I am thinking, thus I be (am)"
– πάντα ῥεῖ
Nov 16 '18 at 4:48
1
@Swordfish Watchworthy: youtube.com/watch?v=29pPZQ77cmI
– πάντα ῥεῖ
Nov 16 '18 at 4:53
1
1
Maybe your forgot "return *this" at the end of op=(...) ?
– Picaud Vincent
Nov 16 '18 at 4:46
Maybe your forgot "return *this" at the end of op=(...) ?
– Picaud Vincent
Nov 16 '18 at 4:46
I thought so too, but it was a copy and paste error. Good catch
– Seoul
Nov 16 '18 at 4:47
I thought so too, but it was a copy and paste error. Good catch
– Seoul
Nov 16 '18 at 4:47
1
1
Not only "*this" but "return *this;"
– Picaud Vincent
Nov 16 '18 at 4:47
Not only "*this" but "return *this;"
– Picaud Vincent
Nov 16 '18 at 4:47
7
7
"Why am I?" Hard to tell, very philosophical question. Some say "I am thinking, thus I be (am)"
– πάντα ῥεῖ
Nov 16 '18 at 4:48
"Why am I?" Hard to tell, very philosophical question. Some say "I am thinking, thus I be (am)"
– πάντα ῥεῖ
Nov 16 '18 at 4:48
1
1
@Swordfish Watchworthy: youtube.com/watch?v=29pPZQ77cmI
– πάντα ῥεῖ
Nov 16 '18 at 4:53
@Swordfish Watchworthy: youtube.com/watch?v=29pPZQ77cmI
– πάντα ῥεῖ
Nov 16 '18 at 4:53
|
show 8 more comments
1 Answer
1
active
oldest
votes
This:
foo& operator=(const auto& f);
will not be standard C++ code until C++20. But gcc has allowed it for quite some time, and what it means is:
template <typename _T>
foo& operator=(const _T& f);
In other words, this is an assignment operator template. It is not a copy assignment operator. That must be a non-template. Since you did not provide a copy assignment operator, the compiler happily generates one for you. In your first code example, you provided your own copy assignment operator.
When you write:
f1 = f2;
In your first example, that expression has one candidate: the copy assignment operator that you wrote. In the second example, there are two candidates: your assignment operator template and the copy assignment operator synthesized by the compiler. The compiler's is a better match (non-template beats template), so it gets called - not yours.
answered Nov 16 '18 at 5:22
BarryBarry
185k21325600
You might want to mention P1141 @chris digged out.
– Swordfish
Nov 16 '18 at 5:27
add a comment |
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1 Answer
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oldest
votes
This:
foo& operator=(const auto& f);
will not be standard C++ code until C++20. But gcc has allowed it for quite some time, and what it means is:
template <typename _T>
foo& operator=(const _T& f);
In other words, this is an assignment operator template. It is not a copy assignment operator. That must be a non-template. Since you did not provide a copy assignment operator, the compiler happily generates one for you. In your first code example, you provided your own copy assignment operator.
When you write:
f1 = f2;
In your first example, that expression has one candidate: the copy assignment operator that you wrote. In the second example, there are two candidates: your assignment operator template and the copy assignment operator synthesized by the compiler. The compiler's is a better match (non-template beats template), so it gets called - not yours.
answered Nov 16 '18 at 5:22
BarryBarry
185k21325600
You might want to mention P1141 @chris digged out.
– Swordfish
Nov 16 '18 at 5:27
add a comment |
This:
foo& operator=(const auto& f);
will not be standard C++ code until C++20. But gcc has allowed it for quite some time, and what it means is:
template <typename _T>
foo& operator=(const _T& f);
In other words, this is an assignment operator template. It is not a copy assignment operator. That must be a non-template. Since you did not provide a copy assignment operator, the compiler happily generates one for you. In your first code example, you provided your own copy assignment operator.
When you write:
f1 = f2;
In your first example, that expression has one candidate: the copy assignment operator that you wrote. In the second example, there are two candidates: your assignment operator template and the copy assignment operator synthesized by the compiler. The compiler's is a better match (non-template beats template), so it gets called - not yours.
answered Nov 16 '18 at 5:22
BarryBarry
185k21325600
You might want to mention P1141 @chris digged out.
– Swordfish
Nov 16 '18 at 5:27
add a comment |
This:
foo& operator=(const auto& f);
will not be standard C++ code until C++20. But gcc has allowed it for quite some time, and what it means is:
template <typename _T>
foo& operator=(const _T& f);
In other words, this is an assignment operator template. It is not a copy assignment operator. That must be a non-template. Since you did not provide a copy assignment operator, the compiler happily generates one for you. In your first code example, you provided your own copy assignment operator.
When you write:
f1 = f2;
In your first example, that expression has one candidate: the copy assignment operator that you wrote. In the second example, there are two candidates: your assignment operator template and the copy assignment operator synthesized by the compiler. The compiler's is a better match (non-template beats template), so it gets called - not yours.
answered Nov 16 '18 at 5:22
BarryBarry
185k21325600
This:
foo& operator=(const auto& f);
will not be standard C++ code until C++20. But gcc has allowed it for quite some time, and what it means is:
template <typename _T>
foo& operator=(const _T& f);
In other words, this is an assignment operator template. It is not a copy assignment operator. That must be a non-template. Since you did not provide a copy assignment operator, the compiler happily generates one for you. In your first code example, you provided your own copy assignment operator.
When you write:
f1 = f2;
In your first example, that expression has one candidate: the copy assignment operator that you wrote. In the second example, there are two candidates: your assignment operator template and the copy assignment operator synthesized by the compiler. The compiler's is a better match (non-template beats template), so it gets called - not yours.
answered Nov 16 '18 at 5:22
BarryBarry
185k21325600
answered Nov 16 '18 at 5:22
BarryBarry
185k21325600
answered Nov 16 '18 at 5:22
BarryBarry
185k21325600
answered Nov 16 '18 at 5:22
BarryBarry
185k21325600
185k21325600
You might want to mention P1141 @chris digged out.
– Swordfish
Nov 16 '18 at 5:27
add a comment |
You might want to mention P1141 @chris digged out.
– Swordfish
Nov 16 '18 at 5:27
You might want to mention P1141 @chris digged out.
– Swordfish
Nov 16 '18 at 5:27
You might want to mention P1141 @chris digged out.
– Swordfish
Nov 16 '18 at 5:27
add a comment |
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1
Maybe your forgot "return *this" at the end of op=(...) ?
– Picaud Vincent
Nov 16 '18 at 4:46
I thought so too, but it was a copy and paste error. Good catch
– Seoul
Nov 16 '18 at 4:47
1
Not only "*this" but "return *this;"
– Picaud Vincent
Nov 16 '18 at 4:47
7
"Why am I?" Hard to tell, very philosophical question. Some say "I am thinking, thus I be (am)"
– πάντα ῥεῖ
Nov 16 '18 at 4:48
1
@Swordfish Watchworthy: youtube.com/watch?v=29pPZQ77cmI
– πάντα ῥεῖ
Nov 16 '18 at 4:53