具有指针的结构 [英] Structure having pointers
问题描述
我有以下结构，
结构样本
{
int i;
char * p;
};
int main（）
{
样本a;
ap = malloc（10）;
样本b;
b = a;
}
现在我觉得浅色的副本已经完成，如果我只在对象上des
那就会有一个悬空的指针。
我如何克服这个问题，因为C结构不支持
函数？
谢谢你提前!!!
Hi,
I have the following Struct,
Struct Sample
{
int i;
char *p;
};
int main()
{
Sample a;
a.p = malloc(10);
Sample b;
b = a;
}
Now i think a shallow copy is done and if i destry only on the object
there would be a dangling pointer.
How do i overcome this problem as C structures don''t support
functions?
Thanks in advance!!!
推荐答案
当你写b = a你复制的内存b中的struct实例a。
这样就可以复制b中a的指针p。现在ap和bp指向
相同的内存。
要制作结构的深度副本，你必须编写一个函数：
struct Sample * SampleCopy（struct Sample * src）
{
struct Sample * b = NULL;
b =（struct Sample * ）malloc（sizeof（struct Sample））;
bi = src.i;
bp =（char *）malloc（sizeof（char）*（strlen（src。 p）+1））;
memcpy（bp，src.p）;
返回b;
}
int main（）
{
样本a;
ap =（char *）malloc（10）;
样本* b;
b = SampleCopy（& a）;
免费（ap）;
free（a）;
printf（"％s"，b-> p）; //应该工作!!
}
我想通过这种方式你可以解决你的问题......如果我已经理解了。
可能我在代码中犯了一些错误..但我现在还没有
参考！
再见
Gio
On 29 Lug，12：20，sam _... @ yahoo.co.in写道：
Hi,
when you write b=a you copy the memory of the struct instance a in b.
In this way you copy the pointer p of a in b. Now a.p and b.p point to
the same memory.
To make a depth copy of the struct you have to write a function:
struct Sample* SampleCopy(struct Sample* src)
{
struct Sample* b=NULL;
b=(struct Sample*)malloc(sizeof(struct Sample));
b.i=src.i;
b.p=(char*)malloc(sizeof(char)*(strlen(src.p)+1));
memcpy(b.p,src.p);
return b;
}
int main()
{
Sample a;
a.p = (char*)malloc(10);
Sample *b;
b = SampleCopy(&a);
free(a.p);
free(a);
printf("%s",b->p); //should work!!
}
I think in this way you may solve your problem... if I''ve understood.
Probably I''ve committed some errors in the code.. but I''ve not a
reference now!
Bye
Gio
On 29 Lug, 12:20, sam_...@yahoo.co.in wrote:
我有以下结构，
结构样本
{
int i;
char * p;
};
int main（）
{
样品a;
ap = malloc（10）;
样品b;
b = a ;
}
现在我认为浅色副本已经完成，如果我仅在对象上进行destry
会有一个悬垂的指针。
我如何克服这个问题，因为C结构不支持
函数？
提前致谢!!!
Hi,
I have the following Struct,
Struct Sample
{
int i;
char *p;
};
int main()
{
Sample a;
a.p = malloc(10);
Sample b;
b = a;
}
Now i think a shallow copy is done and if i destry only on the object
there would be a dangling pointer.
How do i overcome this problem as C structures don''t support
functions?
Thanks in advance!!!
sa * ****@yahoo.co.in 写道：
sa*****@yahoo.co.in wrote:
我有以下结构，
结构样本
Hi,
I have the following Struct,
Struct Sample
关键字是struct，而不是Struct。 C区分大小写。
The keyword is struct, not Struct. C is case sensitive.
{
int i;
char * p;
};
int main（）
{
样本a;
{
int i;
char *p;
};
int main()
{
Sample a;
声明应该是：
struct Sample a;
The declaration should be:
struct Sample a;
ap = malloc（10）;
样本b;
a.p = malloc(10);
Sample b;
如上所述。
As above.
b = a;
}
现在我认为浅拷贝已完成
b = a;
}
Now i think a shallow copy is done
由于ai具有不确定的值，因此副本会调用未定义的行为。
Since a.i has an indeterminate value, the copy invokes undefined behaviour.
如果我只在对象上徘徊
会有一个悬空指针。
如何克服这个问题C结构不支持
功能？
and if i destry only on the object
there would be a dangling pointer.
How do i overcome this problem as C structures don''t support
functions?
通过在ap或bp上调用free释放内存，并将它们设置为
为NULL。
Deallocate the memory by calling free on either a.p or b.p, and set them
both to NULL.
AnticitizenOne写道：
[修正后的帖子]
AnticitizenOne wrote:
[top-post corrected]
On 29 Lug，12：20，sam _... @ yahoo.co.in写道：
On 29 Lug, 12:20, sam_...@yahoo.co.in wrote:
>
我有以下结构，
结构样本
{
int i;
char * p;
};
int main（ ）
{
样本a;
ap = malloc（10）;
样本b;
b = a;
}
现在我认为浅色副本已经完成，如果我只在对象上进行描述，那么会有一个悬空指针。
我如何克服这个问题，因为C结构不会支持
功能？
>Hi,
I have the following Struct,
Struct Sample
{
int i;
char *p;
};
int main()
{
Sample a;
a.p = malloc(10);
Sample b;
b = a;
}
Now i think a shallow copy is done and if i destry only on the object
there would be a dangling pointer.
How do i overcome this problem as C structures don''t support
functions?
当你写b = a时你在b中复制结构实例a的内存。
这样就可以复制b中a的指针p。现在ap和bp指向
相同的内存。
要制作结构的深度副本，你必须编写一个函数：
struct Sample * SampleCopy（struct Sample * src）
{
struct Sample * b = NULL;
b =（struct Sample * ）malloc（sizeof（struct Sample））;
Hi,
when you write b=a you copy the memory of the struct instance a in b.
In this way you copy the pointer p of a in b. Now a.p and b.p point to
the same memory.
To make a depth copy of the struct you have to write a function:
struct Sample* SampleCopy(struct Sample* src)
{
struct Sample* b=NULL;
b=(struct Sample*)malloc(sizeof(struct Sample));
C中不推荐演员。
The cast isn''t recommended in C.
bi = src.i;
bp =（char *）malloc（sizeof（char）*（strlen（src.p）+1））;
b.i=src.i;
b.p=(char*)malloc(sizeof(char)*(strlen(src.p)+1));
而且sizeof（char）在C中总是一个。
And sizeof(char) is always one in C.
memcpy（bp，src.p ）;
返回b;
}
int main（）
{
样本a;
memcpy(b.p,src.p);
return b;
}
int main()
{
Sample a;
它应该是struct Sample a;
It should be struct Sample a;
ap =（char *）malloc（10）;
样本* b;
a.p = (char*)malloc(10);
Sample *b;
C89中不允许使用混合代码和声明，但在
C99中允许它们。
Mixed code and declarations are not allowed in C89 though they''re allowed in
C99.
b = SampleCopy（& a）;
免费（ap）;
免费（a）;
printf（"％s"，b-> p）; //应该管用！！
b = SampleCopy(&a);
free(a.p);
free(a);
printf("%s",b->p); //should work!!
返回0;
return 0;
}
}
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