How to get the returned reference to a vector?

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I found two ways to get the reference returned by the function.

vector<int> vec1 = 4,5,6;

vector<int>& rtn_vec(void)

 return vec1;


int main()

 vector<int> &vec2 = rtn_vec(); //way 1
 vector<int> vec3 = rtn_vec(); //way2

 vec2[0] = 3;

 return 0;

I understand way 1 means passing the reference to vec1 to &vec2, so vec2[0] = 3; changes vec1 to 3,5,6.

But about way 2, I have 2 questions:

Why can I pass a reference (vector<int>&) to an instance (vector<int>), how does it work?

Does way 2 involve deep copy? Because I run this code and vector<int> vec3 = rtn_vec(); seems just copy vec1 to vec3.

edited Nov 15 '18 at 15:16

Toby Speight

17.6k134469

asked Nov 15 '18 at 15:12

Wei-Shou Yang

1

Since the function returns a reference to vec1, way 1 is equivalent to vector<int>& vec2 = vec1;, way 2 is equivalent to vector<int> vec3 = vec1;.

– molbdnilo
Nov 15 '18 at 15:23

add a comment |

I found two ways to get the reference returned by the function.

vector<int> vec1 = 4,5,6;

vector<int>& rtn_vec(void)

 return vec1;


int main()

 vector<int> &vec2 = rtn_vec(); //way 1
 vector<int> vec3 = rtn_vec(); //way2

 vec2[0] = 3;

 return 0;

I understand way 1 means passing the reference to vec1 to &vec2, so vec2[0] = 3; changes vec1 to 3,5,6.

But about way 2, I have 2 questions:

Why can I pass a reference (vector<int>&) to an instance (vector<int>), how does it work?

Does way 2 involve deep copy? Because I run this code and vector<int> vec3 = rtn_vec(); seems just copy vec1 to vec3.

edited Nov 15 '18 at 15:16

Toby Speight

17.6k134469

asked Nov 15 '18 at 15:12

Wei-Shou Yang

1

Since the function returns a reference to vec1, way 1 is equivalent to vector<int>& vec2 = vec1;, way 2 is equivalent to vector<int> vec3 = vec1;.

– molbdnilo
Nov 15 '18 at 15:23

add a comment |

I found two ways to get the reference returned by the function.

vector<int> vec1 = 4,5,6;

vector<int>& rtn_vec(void)

 return vec1;


int main()

 vector<int> &vec2 = rtn_vec(); //way 1
 vector<int> vec3 = rtn_vec(); //way2

 vec2[0] = 3;

 return 0;

I understand way 1 means passing the reference to vec1 to &vec2, so vec2[0] = 3; changes vec1 to 3,5,6.

But about way 2, I have 2 questions:

Why can I pass a reference (vector<int>&) to an instance (vector<int>), how does it work?

Does way 2 involve deep copy? Because I run this code and vector<int> vec3 = rtn_vec(); seems just copy vec1 to vec3.

edited Nov 15 '18 at 15:16

Toby Speight

17.6k134469

asked Nov 15 '18 at 15:12

Wei-Shou Yang

I found two ways to get the reference returned by the function.

vector<int> vec1 = 4,5,6;

vector<int>& rtn_vec(void)

 return vec1;


int main()

 vector<int> &vec2 = rtn_vec(); //way 1
 vector<int> vec3 = rtn_vec(); //way2

 vec2[0] = 3;

 return 0;

I understand way 1 means passing the reference to vec1 to &vec2, so vec2[0] = 3; changes vec1 to 3,5,6.

But about way 2, I have 2 questions:

Why can I pass a reference (vector<int>&) to an instance (vector<int>), how does it work?

Does way 2 involve deep copy? Because I run this code and vector<int> vec3 = rtn_vec(); seems just copy vec1 to vec3.

c++ reference return-value

edited Nov 15 '18 at 15:16

Toby Speight

17.6k134469

asked Nov 15 '18 at 15:12

Wei-Shou Yang

edited Nov 15 '18 at 15:16

Toby Speight

17.6k134469

asked Nov 15 '18 at 15:12

Wei-Shou Yang

edited Nov 15 '18 at 15:16

Toby Speight

17.6k134469

edited Nov 15 '18 at 15:16

Toby Speight

17.6k134469

edited Nov 15 '18 at 15:16

Toby Speight

17.6k134469

asked Nov 15 '18 at 15:12

Wei-Shou Yang

asked Nov 15 '18 at 15:12

Wei-Shou Yang

asked Nov 15 '18 at 15:12

Wei-Shou Yang

1

Since the function returns a reference to vec1, way 1 is equivalent to vector<int>& vec2 = vec1;, way 2 is equivalent to vector<int> vec3 = vec1;.

– molbdnilo
Nov 15 '18 at 15:23

add a comment |

1

Since the function returns a reference to vec1, way 1 is equivalent to vector<int>& vec2 = vec1;, way 2 is equivalent to vector<int> vec3 = vec1;.

– molbdnilo
Nov 15 '18 at 15:23

Since the function returns a reference to vec1, way 1 is equivalent to vector<int>& vec2 = vec1;, way 2 is equivalent to vector<int> vec3 = vec1;.

– molbdnilo
Nov 15 '18 at 15:23

add a comment |

2 Answers
2

active

oldest

votes

vector<int> vec3 = rtn_vec(); //way2

This allocates a new vector and invokes a copy constructor, so yes, this is "deep" copy.

Actually, this is in no way different from simply writing

vector<int> &vec2 = vec1;
vector<int> vec3 = vec1;

Or to make things even clearer

vector<int> &return_value = vec1;
vector<int> &vec2 = return_value;
vector<int> vec3 = return_value;

(Though be careful with term "deep". If it was vector<int*>, then only the pointers would be copied, not the ints themselves.)

edited Nov 16 '18 at 6:32

answered Nov 15 '18 at 15:14

Petr

8,28711843

Thanks for explanation, but I still don't know why can I do vector<int> &return_value = vec1; vector<int> vec3 = return_value; Because return_value is a reference, but vec3 is not a reference. I thought reference is similar to pointer, but I can't do int a; int *p = &a; int b = p; this cause build error, so why can I do this while using reference ?

– Wei-Shou Yang
Nov 17 '18 at 2:30

@Wei-ShouYang, in your example instead of int b = p; you should do int b = *p. References are somewhat like pointers (and technically are pointers), just you don't need to use & for address and * for accessing the value.

– Petr
Nov 17 '18 at 6:26

add a comment |

When you copy-construct vec3, a shallow copy is made (C++ doesn't really do "deep" copy). All the elements in the vector are copied by value, just as with any other copy of a std::vector.

edited Nov 15 '18 at 15:34

answered Nov 15 '18 at 15:17

Toby Speight

17.6k134469

I don't see how you could copy vec and more "deeply" than what happens with vector<int> vec3 = rtn_vec();. It's not a vector of pointers.

– François Andrieux
Nov 15 '18 at 15:31

No, but if it was a vector of pointers, then the pointed-to values wouldn't be copied. Hence, it's a shallow copy.

– Toby Speight
Nov 15 '18 at 15:33

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

vector<int> vec3 = rtn_vec(); //way2

This allocates a new vector and invokes a copy constructor, so yes, this is "deep" copy.

Actually, this is in no way different from simply writing

vector<int> &vec2 = vec1;
vector<int> vec3 = vec1;

Or to make things even clearer

vector<int> &return_value = vec1;
vector<int> &vec2 = return_value;
vector<int> vec3 = return_value;

(Though be careful with term "deep". If it was vector<int*>, then only the pointers would be copied, not the ints themselves.)

edited Nov 16 '18 at 6:32

answered Nov 15 '18 at 15:14

Petr

8,28711843

Thanks for explanation, but I still don't know why can I do vector<int> &return_value = vec1; vector<int> vec3 = return_value; Because return_value is a reference, but vec3 is not a reference. I thought reference is similar to pointer, but I can't do int a; int *p = &a; int b = p; this cause build error, so why can I do this while using reference ?

– Wei-Shou Yang
Nov 17 '18 at 2:30

@Wei-ShouYang, in your example instead of int b = p; you should do int b = *p. References are somewhat like pointers (and technically are pointers), just you don't need to use & for address and * for accessing the value.

– Petr
Nov 17 '18 at 6:26

add a comment |

vector<int> vec3 = rtn_vec(); //way2

This allocates a new vector and invokes a copy constructor, so yes, this is "deep" copy.

Actually, this is in no way different from simply writing

vector<int> &vec2 = vec1;
vector<int> vec3 = vec1;

Or to make things even clearer

vector<int> &return_value = vec1;
vector<int> &vec2 = return_value;
vector<int> vec3 = return_value;

(Though be careful with term "deep". If it was vector<int*>, then only the pointers would be copied, not the ints themselves.)

edited Nov 16 '18 at 6:32

answered Nov 15 '18 at 15:14

Petr

8,28711843

Thanks for explanation, but I still don't know why can I do vector<int> &return_value = vec1; vector<int> vec3 = return_value; Because return_value is a reference, but vec3 is not a reference. I thought reference is similar to pointer, but I can't do int a; int *p = &a; int b = p; this cause build error, so why can I do this while using reference ?

– Wei-Shou Yang
Nov 17 '18 at 2:30

@Wei-ShouYang, in your example instead of int b = p; you should do int b = *p. References are somewhat like pointers (and technically are pointers), just you don't need to use & for address and * for accessing the value.

– Petr
Nov 17 '18 at 6:26

add a comment |

vector<int> vec3 = rtn_vec(); //way2

This allocates a new vector and invokes a copy constructor, so yes, this is "deep" copy.

Actually, this is in no way different from simply writing

vector<int> &vec2 = vec1;
vector<int> vec3 = vec1;

Or to make things even clearer

vector<int> &return_value = vec1;
vector<int> &vec2 = return_value;
vector<int> vec3 = return_value;

(Though be careful with term "deep". If it was vector<int*>, then only the pointers would be copied, not the ints themselves.)

edited Nov 16 '18 at 6:32

answered Nov 15 '18 at 15:14

Petr

8,28711843

vector<int> vec3 = rtn_vec(); //way2

This allocates a new vector and invokes a copy constructor, so yes, this is "deep" copy.

Actually, this is in no way different from simply writing

vector<int> &vec2 = vec1;
vector<int> vec3 = vec1;

Or to make things even clearer

vector<int> &return_value = vec1;
vector<int> &vec2 = return_value;
vector<int> vec3 = return_value;

(Though be careful with term "deep". If it was vector<int*>, then only the pointers would be copied, not the ints themselves.)

edited Nov 16 '18 at 6:32

answered Nov 15 '18 at 15:14

Petr

8,28711843

edited Nov 16 '18 at 6:32

answered Nov 15 '18 at 15:14

Petr

8,28711843

answered Nov 15 '18 at 15:14

Petr

8,28711843

answered Nov 15 '18 at 15:14

Petr

8,28711843

Thanks for explanation, but I still don't know why can I do vector<int> &return_value = vec1; vector<int> vec3 = return_value; Because return_value is a reference, but vec3 is not a reference. I thought reference is similar to pointer, but I can't do int a; int *p = &a; int b = p; this cause build error, so why can I do this while using reference ?

– Wei-Shou Yang
Nov 17 '18 at 2:30

@Wei-ShouYang, in your example instead of int b = p; you should do int b = *p. References are somewhat like pointers (and technically are pointers), just you don't need to use & for address and * for accessing the value.

– Petr
Nov 17 '18 at 6:26

add a comment |

Thanks for explanation, but I still don't know why can I do vector<int> &return_value = vec1; vector<int> vec3 = return_value; Because return_value is a reference, but vec3 is not a reference. I thought reference is similar to pointer, but I can't do int a; int *p = &a; int b = p; this cause build error, so why can I do this while using reference ?

– Wei-Shou Yang
Nov 17 '18 at 2:30

@Wei-ShouYang, in your example instead of int b = p; you should do int b = *p. References are somewhat like pointers (and technically are pointers), just you don't need to use & for address and * for accessing the value.

– Petr
Nov 17 '18 at 6:26

Thanks for explanation, but I still don't know why can I do vector<int> &return_value = vec1; vector<int> vec3 = return_value; Because return_value is a reference, but vec3 is not a reference. I thought reference is similar to pointer, but I can't do int a; int *p = &a; int b = p; this cause build error, so why can I do this while using reference ?

– Wei-Shou Yang
Nov 17 '18 at 2:30

@Wei-ShouYang, in your example instead of int b = p; you should do int b = *p. References are somewhat like pointers (and technically are pointers), just you don't need to use & for address and * for accessing the value.

– Petr
Nov 17 '18 at 6:26

add a comment |

When you copy-construct vec3, a shallow copy is made (C++ doesn't really do "deep" copy). All the elements in the vector are copied by value, just as with any other copy of a std::vector.

edited Nov 15 '18 at 15:34

answered Nov 15 '18 at 15:17

Toby Speight

17.6k134469

I don't see how you could copy vec and more "deeply" than what happens with vector<int> vec3 = rtn_vec();. It's not a vector of pointers.

– François Andrieux
Nov 15 '18 at 15:31

No, but if it was a vector of pointers, then the pointed-to values wouldn't be copied. Hence, it's a shallow copy.

– Toby Speight
Nov 15 '18 at 15:33

add a comment |

When you copy-construct vec3, a shallow copy is made (C++ doesn't really do "deep" copy). All the elements in the vector are copied by value, just as with any other copy of a std::vector.

edited Nov 15 '18 at 15:34

answered Nov 15 '18 at 15:17

Toby Speight

17.6k134469

I don't see how you could copy vec and more "deeply" than what happens with vector<int> vec3 = rtn_vec();. It's not a vector of pointers.

– François Andrieux
Nov 15 '18 at 15:31

No, but if it was a vector of pointers, then the pointed-to values wouldn't be copied. Hence, it's a shallow copy.

– Toby Speight
Nov 15 '18 at 15:33

add a comment |

When you copy-construct vec3, a shallow copy is made (C++ doesn't really do "deep" copy). All the elements in the vector are copied by value, just as with any other copy of a std::vector.

edited Nov 15 '18 at 15:34

answered Nov 15 '18 at 15:17

Toby Speight

17.6k134469

When you copy-construct vec3, a shallow copy is made (C++ doesn't really do "deep" copy). All the elements in the vector are copied by value, just as with any other copy of a std::vector.

edited Nov 15 '18 at 15:34

answered Nov 15 '18 at 15:17

Toby Speight

17.6k134469

edited Nov 15 '18 at 15:34

answered Nov 15 '18 at 15:17

Toby Speight

17.6k134469

answered Nov 15 '18 at 15:17

Toby Speight

17.6k134469

answered Nov 15 '18 at 15:17

Toby Speight

17.6k134469

I don't see how you could copy vec and more "deeply" than what happens with vector<int> vec3 = rtn_vec();. It's not a vector of pointers.

– François Andrieux
Nov 15 '18 at 15:31

No, but if it was a vector of pointers, then the pointed-to values wouldn't be copied. Hence, it's a shallow copy.

– Toby Speight
Nov 15 '18 at 15:33

add a comment |

I don't see how you could copy vec and more "deeply" than what happens with vector<int> vec3 = rtn_vec();. It's not a vector of pointers.

– François Andrieux
Nov 15 '18 at 15:31

No, but if it was a vector of pointers, then the pointed-to values wouldn't be copied. Hence, it's a shallow copy.

– Toby Speight
Nov 15 '18 at 15:33

I don't see how you could copy vec and more "deeply" than what happens with vector<int> vec3 = rtn_vec();. It's not a vector of pointers.

– François Andrieux
Nov 15 '18 at 15:31

No, but if it was a vector of pointers, then the pointed-to values wouldn't be copied. Hence, it's a shallow copy.

– Toby Speight
Nov 15 '18 at 15:33

add a comment |

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