Allocation as default initialization
Say I have a struct (or class) with a dynamic array, its length, and a constructor:
struct array {
int l;
int* t;
array(int length);
};
array::array(int length) {
l=length;
t=new int[l];
}
I assume everything so far is legal: this is how I would write it (although it maybe not be the only way to do things), but I have seen the following code, that somewhat seems to work:
struct array {
int l;
int* t = new int[l];
array(int length);
}
array::array(int length) {
l=length;
}
It looks bad, so I wonder if this works out of sheer luck and undefined behaviors, or if there is some internal rule that makes this code work fine.
c++ constructor
asked 2 hours ago
tarulentarulen
1,747412
add a comment |
Say I have a struct (or class) with a dynamic array, its length, and a constructor:
struct array {
int l;
int* t;
array(int length);
};
array::array(int length) {
l=length;
t=new int[l];
}
I assume everything so far is legal: this is how I would write it (although it maybe not be the only way to do things), but I have seen the following code, that somewhat seems to work:
struct array {
int l;
int* t = new int[l];
array(int length);
}
array::array(int length) {
l=length;
}
It looks bad, so I wonder if this works out of sheer luck and undefined behaviors, or if there is some internal rule that makes this code work fine.
c++ constructor
asked 2 hours ago
tarulentarulen
1,747412
add a comment |
Say I have a struct (or class) with a dynamic array, its length, and a constructor:
struct array {
int l;
int* t;
array(int length);
};
array::array(int length) {
l=length;
t=new int[l];
}
I assume everything so far is legal: this is how I would write it (although it maybe not be the only way to do things), but I have seen the following code, that somewhat seems to work:
struct array {
int l;
int* t = new int[l];
array(int length);
}
array::array(int length) {
l=length;
}
It looks bad, so I wonder if this works out of sheer luck and undefined behaviors, or if there is some internal rule that makes this code work fine.
c++ constructor
asked 2 hours ago
tarulentarulen
1,747412
Say I have a struct (or class) with a dynamic array, its length, and a constructor:
struct array {
int l;
int* t;
array(int length);
};
array::array(int length) {
l=length;
t=new int[l];
}
I assume everything so far is legal: this is how I would write it (although it maybe not be the only way to do things), but I have seen the following code, that somewhat seems to work:
struct array {
int l;
int* t = new int[l];
array(int length);
}
array::array(int length) {
l=length;
}
It looks bad, so I wonder if this works out of sheer luck and undefined behaviors, or if there is some internal rule that makes this code work fine.
c++ constructor
c++ constructor
asked 2 hours ago
tarulentarulen
1,747412
asked 2 hours ago
tarulentarulen
1,747412
asked 2 hours ago
tarulentarulen
1,747412
asked 2 hours ago
tarulentarulen
1,747412
asked 2 hours ago
tarulentarulen
1,747412
1,747412
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
This code is not correct.
int* t = new int[l]; will happen before l=length;, thus reading the uninitialized variable l. Member initializers are handled before the constructor's body runs.
array::array(int length) : l{length} {}
instead would work because l is declared before t.
However, doing this "by hand" is a bad idea to begin with. You should be using std::vector.
answered 2 hours ago
Baum mit AugenBaum mit Augen
40.9k12117152
add a comment |
The 2nd code snippet might have undefined behavior.
The data members are initialized at the order of how they're declared. For class array, when t is initialized l is not initialized yet. For objects with automatic and dynamic storage duration l will be initialized to indeterminate value, then the usage of l (i.e. new int[l]) leads to UB.
Note that l=length; inside the body of the constructor is just assignment; the initialization of data members has been finished before that.
BTW: With member initializer list the 1st code snippet chould be rewritten as
array::array(int length) : l(length), t(new int[l]) {
}
answered 2 hours ago
songyuanyaosongyuanyao
90.5k11171234
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
This code is not correct.
int* t = new int[l]; will happen before l=length;, thus reading the uninitialized variable l. Member initializers are handled before the constructor's body runs.
array::array(int length) : l{length} {}
instead would work because l is declared before t.
However, doing this "by hand" is a bad idea to begin with. You should be using std::vector.
answered 2 hours ago
Baum mit AugenBaum mit Augen
40.9k12117152
add a comment |
This code is not correct.
int* t = new int[l]; will happen before l=length;, thus reading the uninitialized variable l. Member initializers are handled before the constructor's body runs.
array::array(int length) : l{length} {}
instead would work because l is declared before t.
However, doing this "by hand" is a bad idea to begin with. You should be using std::vector.
answered 2 hours ago
Baum mit AugenBaum mit Augen
40.9k12117152
add a comment |
This code is not correct.
int* t = new int[l]; will happen before l=length;, thus reading the uninitialized variable l. Member initializers are handled before the constructor's body runs.
array::array(int length) : l{length} {}
instead would work because l is declared before t.
However, doing this "by hand" is a bad idea to begin with. You should be using std::vector.
answered 2 hours ago
Baum mit AugenBaum mit Augen
40.9k12117152
This code is not correct.
int* t = new int[l]; will happen before l=length;, thus reading the uninitialized variable l. Member initializers are handled before the constructor's body runs.
array::array(int length) : l{length} {}
instead would work because l is declared before t.
However, doing this "by hand" is a bad idea to begin with. You should be using std::vector.
answered 2 hours ago
Baum mit AugenBaum mit Augen
40.9k12117152
answered 2 hours ago
Baum mit AugenBaum mit Augen
40.9k12117152
answered 2 hours ago
Baum mit AugenBaum mit Augen
40.9k12117152
answered 2 hours ago
Baum mit AugenBaum mit Augen
40.9k12117152
40.9k12117152
add a comment |
add a comment |
The 2nd code snippet might have undefined behavior.
The data members are initialized at the order of how they're declared. For class array, when t is initialized l is not initialized yet. For objects with automatic and dynamic storage duration l will be initialized to indeterminate value, then the usage of l (i.e. new int[l]) leads to UB.
Note that l=length; inside the body of the constructor is just assignment; the initialization of data members has been finished before that.
BTW: With member initializer list the 1st code snippet chould be rewritten as
array::array(int length) : l(length), t(new int[l]) {
}
answered 2 hours ago
songyuanyaosongyuanyao
90.5k11171234
add a comment |
The 2nd code snippet might have undefined behavior.
The data members are initialized at the order of how they're declared. For class array, when t is initialized l is not initialized yet. For objects with automatic and dynamic storage duration l will be initialized to indeterminate value, then the usage of l (i.e. new int[l]) leads to UB.
Note that l=length; inside the body of the constructor is just assignment; the initialization of data members has been finished before that.
BTW: With member initializer list the 1st code snippet chould be rewritten as
array::array(int length) : l(length), t(new int[l]) {
}
answered 2 hours ago
songyuanyaosongyuanyao
90.5k11171234
add a comment |
The 2nd code snippet might have undefined behavior.
The data members are initialized at the order of how they're declared. For class array, when t is initialized l is not initialized yet. For objects with automatic and dynamic storage duration l will be initialized to indeterminate value, then the usage of l (i.e. new int[l]) leads to UB.
Note that l=length; inside the body of the constructor is just assignment; the initialization of data members has been finished before that.
BTW: With member initializer list the 1st code snippet chould be rewritten as
array::array(int length) : l(length), t(new int[l]) {
}
answered 2 hours ago
songyuanyaosongyuanyao
90.5k11171234
The 2nd code snippet might have undefined behavior.
The data members are initialized at the order of how they're declared. For class array, when t is initialized l is not initialized yet. For objects with automatic and dynamic storage duration l will be initialized to indeterminate value, then the usage of l (i.e. new int[l]) leads to UB.
Note that l=length; inside the body of the constructor is just assignment; the initialization of data members has been finished before that.
BTW: With member initializer list the 1st code snippet chould be rewritten as
array::array(int length) : l(length), t(new int[l]) {
}
answered 2 hours ago
songyuanyaosongyuanyao
90.5k11171234
edited 2 hours ago
answered 2 hours ago
songyuanyaosongyuanyao
90.5k11171234
answered 2 hours ago
songyuanyaosongyuanyao
90.5k11171234
answered 2 hours ago
songyuanyaosongyuanyao
90.5k11171234
90.5k11171234
add a comment |
add a comment |
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