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 Jan 28 at 10:24
tarulentarulen
1,803513
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 Jan 28 at 10:24
tarulentarulen
1,803513
7
don't use 'l' as a variable name, It's easily confused with the number '1',
– regomodo
Jan 28 at 13:31
@regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuseI,land1. :-)
– Heinzi
Jan 28 at 14:37
5
id rather people read "Clean Code" instead
– regomodo
Jan 28 at 14:54
2
@Heinzi Or maybe stop using single letter variables damnit.
– Bakuriu
Jan 28 at 20:09
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 Jan 28 at 10:24
tarulentarulen
1,803513
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 Jan 28 at 10:24
tarulentarulen
1,803513
asked Jan 28 at 10:24
tarulentarulen
1,803513
asked Jan 28 at 10:24
tarulentarulen
1,803513
asked Jan 28 at 10:24
tarulentarulen
1,803513
asked Jan 28 at 10:24
tarulentarulen
1,803513
1,803513
7
don't use 'l' as a variable name, It's easily confused with the number '1',
– regomodo
Jan 28 at 13:31
@regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuseI,land1. :-)
– Heinzi
Jan 28 at 14:37
5
id rather people read "Clean Code" instead
– regomodo
Jan 28 at 14:54
2
@Heinzi Or maybe stop using single letter variables damnit.
– Bakuriu
Jan 28 at 20:09
add a comment |
7
don't use 'l' as a variable name, It's easily confused with the number '1',
– regomodo
Jan 28 at 13:31
@regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuseI,land1. :-)
– Heinzi
Jan 28 at 14:37
5
id rather people read "Clean Code" instead
– regomodo
Jan 28 at 14:54
2
@Heinzi Or maybe stop using single letter variables damnit.
– Bakuriu
Jan 28 at 20:09
7
7
don't use 'l' as a variable name, It's easily confused with the number '1',
– regomodo
Jan 28 at 13:31
don't use 'l' as a variable name, It's easily confused with the number '1',
– regomodo
Jan 28 at 13:31
@regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuse
I, l and 1. :-)– Heinzi
Jan 28 at 14:37
@regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuse
I, l and 1. :-)– Heinzi
Jan 28 at 14:37
5
5
id rather people read "Clean Code" instead
– regomodo
Jan 28 at 14:54
id rather people read "Clean Code" instead
– regomodo
Jan 28 at 14:54
2
2
@Heinzi Or maybe stop using single letter variables damnit.
– Bakuriu
Jan 28 at 20:09
@Heinzi Or maybe stop using single letter variables damnit.
– Bakuriu
Jan 28 at 20:09
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 Jan 28 at 10:29
Baum mit AugenBaum mit Augen
41.3k12118155
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 Jan 28 at 10:28
songyuanyaosongyuanyao
92.5k11178243
3
Why do you say "might"? It does have UB, unconditionally.
– Ruslan
Jan 28 at 16:29
1
@Ruslan Because for static or thread-local objectslwill get zero initialized at first.
– songyuanyao
Jan 29 at 1:13
@Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)
– Lightness Races in Orbit
Jan 30 at 16:33
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 Jan 28 at 10:29
Baum mit AugenBaum mit Augen
41.3k12118155
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 Jan 28 at 10:29
Baum mit AugenBaum mit Augen
41.3k12118155
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 Jan 28 at 10:29
Baum mit AugenBaum mit Augen
41.3k12118155
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 Jan 28 at 10:29
Baum mit AugenBaum mit Augen
41.3k12118155
answered Jan 28 at 10:29
Baum mit AugenBaum mit Augen
41.3k12118155
answered Jan 28 at 10:29
Baum mit AugenBaum mit Augen
41.3k12118155
answered Jan 28 at 10:29
Baum mit AugenBaum mit Augen
41.3k12118155
41.3k12118155
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 Jan 28 at 10:28
songyuanyaosongyuanyao
92.5k11178243
3
Why do you say "might"? It does have UB, unconditionally.
– Ruslan
Jan 28 at 16:29
1
@Ruslan Because for static or thread-local objectslwill get zero initialized at first.
– songyuanyao
Jan 29 at 1:13
@Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)
– Lightness Races in Orbit
Jan 30 at 16:33
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 Jan 28 at 10:28
songyuanyaosongyuanyao
92.5k11178243
3
Why do you say "might"? It does have UB, unconditionally.
– Ruslan
Jan 28 at 16:29
1
@Ruslan Because for static or thread-local objectslwill get zero initialized at first.
– songyuanyao
Jan 29 at 1:13
@Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)
– Lightness Races in Orbit
Jan 30 at 16:33
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 Jan 28 at 10:28
songyuanyaosongyuanyao
92.5k11178243
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 Jan 28 at 10:28
songyuanyaosongyuanyao
92.5k11178243
edited Jan 28 at 10:33
answered Jan 28 at 10:28
songyuanyaosongyuanyao
92.5k11178243
answered Jan 28 at 10:28
songyuanyaosongyuanyao
92.5k11178243
answered Jan 28 at 10:28
songyuanyaosongyuanyao
92.5k11178243
92.5k11178243
3
Why do you say "might"? It does have UB, unconditionally.
– Ruslan
Jan 28 at 16:29
1
@Ruslan Because for static or thread-local objectslwill get zero initialized at first.
– songyuanyao
Jan 29 at 1:13
@Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)
– Lightness Races in Orbit
Jan 30 at 16:33
add a comment |
3
Why do you say "might"? It does have UB, unconditionally.
– Ruslan
Jan 28 at 16:29
1
@Ruslan Because for static or thread-local objectslwill get zero initialized at first.
– songyuanyao
Jan 29 at 1:13
@Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)
– Lightness Races in Orbit
Jan 30 at 16:33
3
3
Why do you say "might"? It does have UB, unconditionally.
– Ruslan
Jan 28 at 16:29
Why do you say "might"? It does have UB, unconditionally.
– Ruslan
Jan 28 at 16:29
1
1
@Ruslan Because for static or thread-local objects
l will get zero initialized at first.– songyuanyao
Jan 29 at 1:13
@Ruslan Because for static or thread-local objects
l will get zero initialized at first.– songyuanyao
Jan 29 at 1:13
@Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)
– Lightness Races in Orbit
Jan 30 at 16:33
@Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)
– Lightness Races in Orbit
Jan 30 at 16:33
add a comment |
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7
don't use 'l' as a variable name, It's easily confused with the number '1',
– regomodo
Jan 28 at 13:31
@regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuse
I,land1. :-)– Heinzi
Jan 28 at 14:37
5
id rather people read "Clean Code" instead
– regomodo
Jan 28 at 14:54
2
@Heinzi Or maybe stop using single letter variables damnit.
– Bakuriu
Jan 28 at 20:09