Mixing
Returning a variable about to be destroyed in a function
j is destroyed when the function calls return
k is destroyed at the end of the enclosing brackets
If i pass in 9 for j, k is created and will be assigned 81
Returning k will set func1 which is a reference to an integer = k
Returning will immediately terminate the function
My question is, are k and j terminated at the return statement?
If they are func1 should reference nothing...
But i have tried to run this code and it works...
int& func1(int j){
int k = j*j;
return(k);
}
c++
asked Nov 19 '18 at 13:20
Jokaaa
124
|
show 3 more comments
j is destroyed when the function calls return
k is destroyed at the end of the enclosing brackets
If i pass in 9 for j, k is created and will be assigned 81
Returning k will set func1 which is a reference to an integer = k
Returning will immediately terminate the function
My question is, are k and j terminated at the return statement?
If they are func1 should reference nothing...
But i have tried to run this code and it works...
int& func1(int j){
int k = j*j;
return(k);
}
c++
asked Nov 19 '18 at 13:20
Jokaaa
124
Possible duplicate of C++ Returning reference to local variable
– Algirdas Preidžius
Nov 19 '18 at 13:24
1
"But i have tried to run this code and it works..." "It works" is only one of (many) possible manifestations of undefined behavior. It may even continue to work, up until the point you decide to show your code to a potential client, when it might decide to stop working.
– Algirdas Preidžius
Nov 19 '18 at 13:26
This should also trigger a warning from the compiler depending on your compiler and warning level.
– uceumern
Nov 19 '18 at 13:50
@Algirdas Preidžius OMG you are so smart, can we be friends??
– Jokaaa
Nov 19 '18 at 14:31
What is the reason for it continuing to work, compiler had a memory lapse?
– Jokaaa
Nov 19 '18 at 14:33
|
show 3 more comments
j is destroyed when the function calls return
k is destroyed at the end of the enclosing brackets
If i pass in 9 for j, k is created and will be assigned 81
Returning k will set func1 which is a reference to an integer = k
Returning will immediately terminate the function
My question is, are k and j terminated at the return statement?
If they are func1 should reference nothing...
But i have tried to run this code and it works...
int& func1(int j){
int k = j*j;
return(k);
}
c++
asked Nov 19 '18 at 13:20
Jokaaa
124
j is destroyed when the function calls return
k is destroyed at the end of the enclosing brackets
If i pass in 9 for j, k is created and will be assigned 81
Returning k will set func1 which is a reference to an integer = k
Returning will immediately terminate the function
My question is, are k and j terminated at the return statement?
If they are func1 should reference nothing...
But i have tried to run this code and it works...
int& func1(int j){
int k = j*j;
return(k);
}
c++
c++
asked Nov 19 '18 at 13:20
Jokaaa
124
asked Nov 19 '18 at 13:20
Jokaaa
124
asked Nov 19 '18 at 13:20
Jokaaa
124
asked Nov 19 '18 at 13:20
Jokaaa
124
asked Nov 19 '18 at 13:20
Jokaaa
124
124
Possible duplicate of C++ Returning reference to local variable
– Algirdas Preidžius
Nov 19 '18 at 13:24
1
"But i have tried to run this code and it works..." "It works" is only one of (many) possible manifestations of undefined behavior. It may even continue to work, up until the point you decide to show your code to a potential client, when it might decide to stop working.
– Algirdas Preidžius
Nov 19 '18 at 13:26
This should also trigger a warning from the compiler depending on your compiler and warning level.
– uceumern
Nov 19 '18 at 13:50
@Algirdas Preidžius OMG you are so smart, can we be friends??
– Jokaaa
Nov 19 '18 at 14:31
What is the reason for it continuing to work, compiler had a memory lapse?
– Jokaaa
Nov 19 '18 at 14:33
|
show 3 more comments
Possible duplicate of C++ Returning reference to local variable
– Algirdas Preidžius
Nov 19 '18 at 13:24
1
"But i have tried to run this code and it works..." "It works" is only one of (many) possible manifestations of undefined behavior. It may even continue to work, up until the point you decide to show your code to a potential client, when it might decide to stop working.
– Algirdas Preidžius
Nov 19 '18 at 13:26
This should also trigger a warning from the compiler depending on your compiler and warning level.
– uceumern
Nov 19 '18 at 13:50
@Algirdas Preidžius OMG you are so smart, can we be friends??
– Jokaaa
Nov 19 '18 at 14:31
What is the reason for it continuing to work, compiler had a memory lapse?
– Jokaaa
Nov 19 '18 at 14:33
Possible duplicate of C++ Returning reference to local variable
– Algirdas Preidžius
Nov 19 '18 at 13:24
Possible duplicate of C++ Returning reference to local variable
– Algirdas Preidžius
Nov 19 '18 at 13:24
1
1
"But i have tried to run this code and it works..." "It works" is only one of (many) possible manifestations of undefined behavior. It may even continue to work, up until the point you decide to show your code to a potential client, when it might decide to stop working.
– Algirdas Preidžius
Nov 19 '18 at 13:26
"But i have tried to run this code and it works..." "It works" is only one of (many) possible manifestations of undefined behavior. It may even continue to work, up until the point you decide to show your code to a potential client, when it might decide to stop working.
– Algirdas Preidžius
Nov 19 '18 at 13:26
This should also trigger a warning from the compiler depending on your compiler and warning level.
– uceumern
Nov 19 '18 at 13:50
This should also trigger a warning from the compiler depending on your compiler and warning level.
– uceumern
Nov 19 '18 at 13:50
@Algirdas Preidžius OMG you are so smart, can we be friends??
– Jokaaa
Nov 19 '18 at 14:31
@Algirdas Preidžius OMG you are so smart, can we be friends??
– Jokaaa
Nov 19 '18 at 14:31
What is the reason for it continuing to work, compiler had a memory lapse?
– Jokaaa
Nov 19 '18 at 14:33
What is the reason for it continuing to work, compiler had a memory lapse?
– Jokaaa
Nov 19 '18 at 14:33
|
show 3 more comments
2 Answers
2
active
oldest
votes
and it works...
No, it appears to work. The moment you try to access the reference returned by func1 you enter the Undefined Behavior realm. At that point all bets are off, it could've printed out 42, printed out nothing, crash, eat your CMOS battery etc.
answered Nov 19 '18 at 13:30
Sombrero Chicken
23.3k33077
Nice name, how do you come up with it
– Jokaaa
Nov 19 '18 at 14:29
add a comment |
If they are func1 should reference nothing...
You are correct. Try using the reference some more time and you will see:
#include <iostream>
int& func1(int j) {
int k = j * j;
return(k);
}
int main() {
int& addr = func1(9);
for (int i = 0; i < 10; ++i) {
std::cout << addr << 'n';
}
}
Output:
81
2758456
2758456
2758456
2758456
2758456
2758456
2758456
2758456
2758456
answered Nov 19 '18 at 13:26
Stack Danny
1,105319
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
and it works...
No, it appears to work. The moment you try to access the reference returned by func1 you enter the Undefined Behavior realm. At that point all bets are off, it could've printed out 42, printed out nothing, crash, eat your CMOS battery etc.
answered Nov 19 '18 at 13:30
Sombrero Chicken
23.3k33077
Nice name, how do you come up with it
– Jokaaa
Nov 19 '18 at 14:29
add a comment |
and it works...
No, it appears to work. The moment you try to access the reference returned by func1 you enter the Undefined Behavior realm. At that point all bets are off, it could've printed out 42, printed out nothing, crash, eat your CMOS battery etc.
answered Nov 19 '18 at 13:30
Sombrero Chicken
23.3k33077
Nice name, how do you come up with it
– Jokaaa
Nov 19 '18 at 14:29
add a comment |
and it works...
No, it appears to work. The moment you try to access the reference returned by func1 you enter the Undefined Behavior realm. At that point all bets are off, it could've printed out 42, printed out nothing, crash, eat your CMOS battery etc.
answered Nov 19 '18 at 13:30
Sombrero Chicken
23.3k33077
and it works...
No, it appears to work. The moment you try to access the reference returned by func1 you enter the Undefined Behavior realm. At that point all bets are off, it could've printed out 42, printed out nothing, crash, eat your CMOS battery etc.
answered Nov 19 '18 at 13:30
Sombrero Chicken
23.3k33077
answered Nov 19 '18 at 13:30
Sombrero Chicken
23.3k33077
answered Nov 19 '18 at 13:30
Sombrero Chicken
23.3k33077
answered Nov 19 '18 at 13:30
Sombrero Chicken
23.3k33077
23.3k33077
Nice name, how do you come up with it
– Jokaaa
Nov 19 '18 at 14:29
add a comment |
Nice name, how do you come up with it
– Jokaaa
Nov 19 '18 at 14:29
Nice name, how do you come up with it
– Jokaaa
Nov 19 '18 at 14:29
Nice name, how do you come up with it
– Jokaaa
Nov 19 '18 at 14:29
add a comment |
If they are func1 should reference nothing...
You are correct. Try using the reference some more time and you will see:
#include <iostream>
int& func1(int j) {
int k = j * j;
return(k);
}
int main() {
int& addr = func1(9);
for (int i = 0; i < 10; ++i) {
std::cout << addr << 'n';
}
}
Output:
81
2758456
2758456
2758456
2758456
2758456
2758456
2758456
2758456
2758456
answered Nov 19 '18 at 13:26
Stack Danny
1,105319
add a comment |
If they are func1 should reference nothing...
You are correct. Try using the reference some more time and you will see:
#include <iostream>
int& func1(int j) {
int k = j * j;
return(k);
}
int main() {
int& addr = func1(9);
for (int i = 0; i < 10; ++i) {
std::cout << addr << 'n';
}
}
Output:
81
2758456
2758456
2758456
2758456
2758456
2758456
2758456
2758456
2758456
answered Nov 19 '18 at 13:26
Stack Danny
1,105319
add a comment |
If they are func1 should reference nothing...
You are correct. Try using the reference some more time and you will see:
#include <iostream>
int& func1(int j) {
int k = j * j;
return(k);
}
int main() {
int& addr = func1(9);
for (int i = 0; i < 10; ++i) {
std::cout << addr << 'n';
}
}
Output:
81
2758456
2758456
2758456
2758456
2758456
2758456
2758456
2758456
2758456
answered Nov 19 '18 at 13:26
Stack Danny
1,105319
If they are func1 should reference nothing...
You are correct. Try using the reference some more time and you will see:
#include <iostream>
int& func1(int j) {
int k = j * j;
return(k);
}
int main() {
int& addr = func1(9);
for (int i = 0; i < 10; ++i) {
std::cout << addr << 'n';
}
}
Output:
81
2758456
2758456
2758456
2758456
2758456
2758456
2758456
2758456
2758456
answered Nov 19 '18 at 13:26
Stack Danny
1,105319
answered Nov 19 '18 at 13:26
Stack Danny
1,105319
answered Nov 19 '18 at 13:26
Stack Danny
1,105319
answered Nov 19 '18 at 13:26
Stack Danny
1,105319
1,105319
add a comment |
add a comment |
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Possible duplicate of C++ Returning reference to local variable
– Algirdas Preidžius
Nov 19 '18 at 13:24
1
"But i have tried to run this code and it works..." "It works" is only one of (many) possible manifestations of undefined behavior. It may even continue to work, up until the point you decide to show your code to a potential client, when it might decide to stop working.
– Algirdas Preidžius
Nov 19 '18 at 13:26
This should also trigger a warning from the compiler depending on your compiler and warning level.
– uceumern
Nov 19 '18 at 13:50
@Algirdas Preidžius OMG you are so smart, can we be friends??
– Jokaaa
Nov 19 '18 at 14:31
What is the reason for it continuing to work, compiler had a memory lapse?
– Jokaaa
Nov 19 '18 at 14:33