为指针分配内存 [英] allocate memory for pointer
问题描述
我遇到了分段错误我检查了我的代码，发现
以下问题：
我想为数组重新分配内存。
我定义了以下函数：
int reallocateMemory（double * array，int newsize）
{
if（array）delete []数组;
array = new double [newsize];
返回1;
}
现在，
int main（）
{
double * a = new double [2]，* b = new double [10];
cout<< a<< ENDL; //的地址（1）
for（int i = 0; i< 10; i ++）b [i] = i; // b = {0,1，....，9}
reallocateMemory（a，10）;
cout< < a<< ENDL; // a的地址与（1）相同！ :(
for（int i = 0; i< 10; i ++）a [i] = 0.1 * i;
for （int i = 0; i< 10; i ++）
cout<< a [i]<<"" ;;
cout< < endl; //很好，a是0,0.1，...，0.9
for（int i = 0; i< 10; i ++）
cout<< b [i]<<"";
cout<< endl;
//错误：输出= 0.3,0.4，...，0.9,7,8,9
返回0;
}
所以a的内存没有被重新分配，虽然我把这个函数称为
更改它。
我认为是什么函数中的pass-by-pointer将直接更改
参数。因此，在我调用函数
reallocateMemory之后，应该更改a。但似乎是结果是
与传值相同。
我的功能有什么问题？
感谢您的帮助！
X
I encountered "segmentation fault" and I checked my code, found the
following problem:
I want to reallocate memory for an array.
I defined the following function:
int reallocateMemory( double *array, int newsize )
{
if (array) delete[] array;
array = new double[newsize];
return 1;
}
Now,
int main()
{
double *a = new double[2], *b = new double[10];
cout << a << endl; // address of a. (1)
for ( int i = 0; i < 10; i++ ) b[i] = i; // b = {0,1,....,9}
reallocateMemory( a, 10 );
cout << a << endl; // the address of a is same as (1)! :(
for ( int i = 0; i < 10; i++ ) a[i] = 0.1*i;
for ( int i = 0; i < 10; i++ )
cout << a[i] << " ";
cout << endl; // fine, a is 0,0.1,...,0.9
for ( int i = 0; i < 10; i++ )
cout << b[i] << " ";
cout << endl;
// error: output = 0.3,0.4,...,0.9,7,8,9
return 0;
}
So the memory of "a" is not reallocated although I call the function to
change it.
What I think is that passed-by-pointer in function will change the
parameter directly. Therefore after I call the function
reallocateMemory, "a" should be changed. But it seems that the result is
same as passed-by-value.
What''s the problem in my function?
Thank you for your help!
X
推荐答案
" xuatla" <徐**** @ gmail.com>在消息中写道
news：40 ************** @ gmail.com ...
"xuatla" <xu****@gmail.com> wrote in message
news:40**************@gmail.com...
我遇到了分段错误我检查了我的代码，发现了以下问题：
我想为数组重新分配内存。
我定义了以下功能：
int reallocateMemory（double * array，int newsize）
{
if（array）delete [] array;
array = new double [newsize];
如何将阵列返回到主？
返回1;
}
现在，
int main（）
{
double * a = new double [2]，* b = new double [10];
cout<< a<< ENDL; //的地址（1）
for（int i = 0; i< 10; i ++）b [i] = i; // b = {0,1，....，9}
reallocateMemory（a，10）;
I encountered "segmentation fault" and I checked my code, found the
following problem:
I want to reallocate memory for an array.
I defined the following function:
int reallocateMemory( double *array, int newsize )
{
if (array) delete[] array;
array = new double[newsize];
How do you return array back to main now?
return 1;
}
Now,
int main()
{
double *a = new double[2], *b = new double[10];
cout << a << endl; // address of a. (1)
for ( int i = 0; i < 10; i++ ) b[i] = i; // b = {0,1,....,9}
reallocateMemory( a, 10 );
此函数后a指向已删除的内存。从此尝试任何东西
在那个内存上是UB。
所以它解决了问题变更重新分配到 -
int reallocateMemory（double *& array，int newsize）
^^^^^^^^
您现在正在调用对指针的引用。现在对阵列
所做的任何更改都将反映在主要内容中。
-Sharad
After this function a points to deleted memory. Henceforth trying anything
on that memory is UB.
So it resolve the problem change reallocate to -
int reallocateMemory( double *&array, int newsize )
^^^^^^^^
You are now paasing a reference to a pointer. Any changes now made to array
will be reflected back in main.
-Sharad
2004年7月15日星期四00:20:39 -0700，xuatla< xu **** @ gmail.com>写道：
On Thu, 15 Jul 2004 00:20:39 -0700, xuatla <xu****@gmail.com> wrote:
我遇到了分段错误我检查了我的代码，发现了以下问题：
我想为数组重新分配内存。
我定义了以下功能：
int reallocateMemory（double * array，int newsize）
{
if（array）delete [] array;
array = new double [newsize];
返回1;
}
现在，
int main（）
{
double * a = new double [2]，* b =新双[10];
cout<< a<< ENDL; //的地址（1）
for（int i = 0; i< 10; i ++）b [i] = i; // b = {0,1，....，9}
reallocateMemory（a，10）;
cout<< a<< ENDL; // a的地址与（1）相同！ :( int i = 0; i< 10; i ++）a [i] = 0.1 * i;
for（int i = 0; i< ; 10; i ++）
cout<< a [i]<<"";
cout<< endl; //罚款，a是0,0.1，.. （int i = 0; i< 10; i ++）
cout<< b [i]<<"";
cout<< endl;
//错误：输出= 0.3,0.4，...，0.9,7,8,9
返回0;
}
所以a的记忆并没有重新分配，虽然我把这个功能叫做改变它。
我认为通过了 - 函数中的by-pointer将直接改变
参数。因此，在我调用函数
reallocateMemory后，应更改a。但似乎结果与通过的相同 - by-value。
是的，指针和其他一切一样按值传递。
我的函数有什么问题？
I encountered "segmentation fault" and I checked my code, found the
following problem:
I want to reallocate memory for an array.
I defined the following function:
int reallocateMemory( double *array, int newsize )
{
if (array) delete[] array;
array = new double[newsize];
return 1;
}
Now,
int main()
{
double *a = new double[2], *b = new double[10];
cout << a << endl; // address of a. (1)
for ( int i = 0; i < 10; i++ ) b[i] = i; // b = {0,1,....,9}
reallocateMemory( a, 10 );
cout << a << endl; // the address of a is same as (1)! :(
for ( int i = 0; i < 10; i++ ) a[i] = 0.1*i;
for ( int i = 0; i < 10; i++ )
cout << a[i] << " ";
cout << endl; // fine, a is 0,0.1,...,0.9
for ( int i = 0; i < 10; i++ )
cout << b[i] << " ";
cout << endl;
// error: output = 0.3,0.4,...,0.9,7,8,9
return 0;
}
So the memory of "a" is not reallocated although I call the function to
change it.
What I think is that passed-by-pointer in function will change the
parameter directly. Therefore after I call the function
reallocateMemory, "a" should be changed. But it seems that the result is
same as passed-by-value.
Yes, pointers are passed by value just like everything else.
What''s the problem in my function?
你正在经过价值。如果你想要o使用一个函数来改变你有三个选择的变量值。
1）使用返回值
double * reallocateMemory（double * array，int newsize）
{
delete [] array;
返回new double [newsize ];
}
a = reallocateMemory（a，10）;
2）使用参考
void reallocateMemory（double *& array，int newsize）
{
delete [] array;
array = new double [newsize];
}
reallocateMemory（a，10）;
3）使用指针（在你的情况下，这将是指向指针的指针） ）
void reallocateMemory（double ** array，int newsize）
{
delete [] * array;
* array = new double [newsize];
}
reallocateMemory（& a，10）;
BTW没有必要在删除指针之前测试NULL
if（ptr）delete [] ptr;
完全正常工作与
删除[] ptr;
删除NULL保证无效。
john
You are passing by value. If you want to use a function to change that
value of a variable you have three alternatives.
1) Use a return value
double* reallocateMemory( double *array, int newsize )
{
delete[] array;
return new double[newsize];
}
a = reallocateMemory( a, 10 );
2) Use a reference
void reallocateMemory( double*& array, int newsize )
{
delete[] array;
array = new double[newsize];
}
reallocateMemory( a, 10 );
3) Use a pointer (in your case this would be a pointer to a pointer)
void reallocateMemory( double** array, int newsize )
{
delete[] *array;
*array = new double[newsize];
}
reallocateMemory( &a, 10 );
BTW it is not necessary to test for NULL before deleteing a pointer
if (ptr) delete[] ptr;
works exactly the same as
delete[] ptr;
Deleting NULL is guaranteed to have no effect.
john
[...]
int reallocateMemory（double * array，int newsize）
{
if（array）delete [] array;
array = new double [newsize];
返回1;
}
[ ...]
我的函数有什么问题？
int reallocateMemory( double *array, int newsize )
{
if (array) delete[] array;
array = new double[newsize];
return 1;
} [...]
What''s the problem in my function?
你的函数只改变了没有
影响的局部变量数组你的变量a在main中。将函数声明更改为
int reallocateMemory（double *& array，int newsize）
应该可以解决问题。现在不是a的值传递给函数
但它的引用因此你可以修改它的值...
your function changes only the local variable array which has no
influence to your variable a in main. Changing the functiondeclaration to
int reallocateMemory( double *&array, int newsize )
should do the trick. Now not the value of a is passed to the function
but the reference of it so you''re able to modify its value...
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