结构和动态内存分配 [英] structs and dynamic memory allocation
问题描述
你好，
我对结构有疑问。我有一个struct（profil）
有一个指向另一个struct（point）的指针。 struct profil存储
坐标点。
问题是我不知道会有多少积分
最后的每一个结构，所以我必须动态地分配内存
，不幸的是不能使用固定大小的数组。
我想知道是否有更好的方法来访问struct成员（
分）。
使用p1->访问struct成员; p2-> x看起来相当丑陋且不安全。
下面是该程序的精简版，只是为了了解什么
我'我想做。我知道这不是一个纯粹的ANSI C定义
问题，但我无法在这个特殊问题中找到任何帮助
K& R也不在FAQ中。所以我赞成每一个提示。
最好的问候
罗马
#include< stdlib.h>
#include< stdio.h>
结构点
{
双x;
双y;
};
struct profil
{
struct point * pt;
struct profil * previous;
struct profil * next;
};
int main（无效）
{
int n;
struct profil * p1;
struct profil * begin;
n = 2; / *仅为此示例分配内存* /
if（（p1 =（struct profil *）malloc（sizeof（struct profil）））== NULL）
{
printf（&\\; \ nMemory allocation failed\\\
"）;
返回1;
}
begin = p1;
if（（p1-> pt =（struct point *）malloc（n * sizeof（struct point）） ）== NULL）
{
printf（&\\; \ nMemory allocation failed\\\
）;
返回1;
}
/ *将一些信息放入结构* /
p1-> pt-> x = 1265.75;
p1-> pt-> y = 150.73;
p1-> pt ++; / *丑陋，我知道。有一个更好的方法吗？ * /
p1-> pt-> x = 550.55;
p1-> pt-> y = 350.33;
p1-> pt--;
/ *用结构* /
/ *代替打印结果，我打印结构* /
printf（" \ nStrukt 1：％f \ n"，p1-> pt-> x）;
printf（" \ nStrukt 1：％f \ n"，p1-> pt-> y）;
p1-> pt ++ ;
printf（" \ nStrukt 2：％f \ n"，p1-> pt-> x）;
printf（ \ nStrukt 2：％f \ n，p1-> pt-> y）;
返回0;
} < br>
hello,
I do have a question regarding structs. I have a struct (profil) which
has a pointer to another struct (point). The struct profil stores the
coordinates of points.
The problem is that I don''t know how many points
there will be in every struct in the end, so I have to allocate memory
dynamically for
them and can''t use an array of fixed size, unfortunately.
I would like to know if there is a better way to access struct members (the
points).
Accessing struct members with p1->p2->x seems rather ugly and unsafe.
Below is a stripped down version of the program, just to get an idea what
I''m trying to do. I''m aware that this is not a pure ANSI C definition
question, but I couldn''t find any help to this special problem neither in
K&R nor in the FAQ. So I apreciate every hint.
Best Regards
Roman
#include <stdlib.h>
#include <stdio.h>
/* struct with points, used by struct profil */
struct point
{
double x;
double y;
};
struct profil
{
struct point *pt;
struct profil *previous;
struct profil *next;
};
int main(void)
{
int n;
struct profil *p1;
struct profil *begin;
n=2; /* only to allocate memory for this example */
if((p1=(struct profil *) malloc(sizeof(struct profil)))==NULL)
{
printf("\nMemory allocation failed\n");
return 1;
}
begin=p1;
if((p1->pt=(struct point *) malloc(n*sizeof(struct point)))==NULL)
{
printf("\nMemory allocation failed\n");
return 1;
}
/* put some information into the struct */
p1->pt->x=1265.75;
p1->pt->y=150.73;
p1->pt++; /* ugly, I know. Is there a better way to do this? */
p1->pt->x=550.55;
p1->pt->y=350.33;
p1->pt--;
/* do something with the structs */
/* instead of printing results, I print the structs out */
printf("\nStrukt 1: %f\n", p1->pt->x);
printf("\nStrukt 1: %f\n", p1->pt->y);
p1->pt++;
printf("\nStrukt 2: %f\n", p1->pt->x);
printf("\nStrukt 2: %f\n", p1->pt->y);
return 0;
}
推荐答案
Roman Hartmann写道：
Roman Hartmann wrote:
我想知道是否有更好的方法来访问struct成员
（要点）。
使用p1-> p2-> x访问struct成员似乎相当丑陋且不安全。
[。 。 。
/ * struct with points，struct profil * /
struct point
{
double x;
double y;
};
struct profil
{struct struct * pt;
struct profil * previous;
struct profil * next;
} ;
int main（无效）
{
int n;
struct profil * p1;
struct profil * begin;
[。 。 。]
/ *将一些信息放入结构中* /
p1-> pt-> x = 1265.75;
p1-> pt-> y = 150.73;
p1-> pt ++; / *丑陋，我知道。有一个更好的方法吗？ * /
是的，这个怎么样：
struct point * pt = p1-> pt;
pt [0] .x = 1265.75;
pt [0] .y = 150.73;
p1-> pt-> x = 550.55;
p1-> pt-> y = 350.33;
p1-> pt--;
pt [1] .x = 550.55;
pt [1] .y = 350.33;
/ *做点什么使用结构* /
/ *而不是打印结果，我打印出结构* /
printf（" \ nStrukt 1：％f \ n，p1-> ; pt-> x）;
printf（" \ nStrukt 1：％f \ n"，p1-> pt-> y）;
p1-> ; pt ++;
printf（" \ nStrukt 2：％f \ n"，p1-> pt-> x）;
printf（" \ nStrukt 2 ：％f \ n"，p1-> pt-> y）;
返回0;
}
I would like to know if there is a better way to access struct members
(the points).
Accessing struct members with p1->p2->x seems rather ugly and unsafe.
[. . .]
/* struct with points, used by struct profil */
struct point
{
double x;
double y;
};
struct profil
{
struct point *pt;
struct profil *previous;
struct profil *next;
};
int main(void)
{
int n;
struct profil *p1;
struct profil *begin;
[. . .]
/* put some information into the struct */
p1->pt->x=1265.75;
p1->pt->y=150.73;
p1->pt++; /* ugly, I know. Is there a better way to do this? */
Yes, how about this:
struct point *pt = p1->pt;
pt[0].x = 1265.75;
pt[0].y = 150.73;
p1->pt->x=550.55;
p1->pt->y=350.33;
p1->pt--;
pt[1].x = 550.55;
pt[1].y = 350.33;
/* do something with the structs */
/* instead of printing results, I print the structs out */
printf("\nStrukt 1: %f\n", p1->pt->x);
printf("\nStrukt 1: %f\n", p1->pt->y);
p1->pt++;
printf("\nStrukt 2: %f\n", p1->pt->x);
printf("\nStrukt 2: %f\n", p1->pt->y);
return 0;
}
int i;
for（i = 0; i< n; ++ i）
{
printf（" \ nStrukt％d：％f \ n"，i + 1，pt [i] .x）;
printf（" \ nStrukt％d：％f \ n"，i + 1，pt [i] .y）;
}
问候，
Russell Hanneken rg********@pobox.com
删除'g''从我的地址到给我发邮件。
int i;
for (i = 0; i < n; ++i)
{
printf("\nStrukt %d: %f\n", i + 1, pt[i].x);
printf("\nStrukt %d: %f\n", i + 1, pt[i].y);
}
Regards,
Russell Hanneken
rg********@pobox.com
Remove the ''g'' from my address to send me mail.
Russell Hanneken写道：
Russell Hanneken wrote:
Roman Hartmann写道：
Roman Hartmann wrote:
int main（ void）
结构profil * p1;
struct profil * begin;
int main(void)
{
int n;
struct profil *p1;
struct profil *begin;
[。 。 。 ]
struct point * pt = p1-> pt;
pt [0] .x = 1265.75;
pt [0] .y = 150.73;
pt [1] .x = 550.55;
pt [1] .y = 350.33;
int i;
[ . . . ]
struct point *pt = p1->pt;
pt[0].x = 1265.75;
pt[0].y = 150.73;
pt[1].x = 550.55;
pt[1].y = 350.33;
int i;
哎呀; i和pt都需要在main的开头声明。除非
你编写的代码符合新的（1999）C标准（你可能不是b $ b），所以局部变量声明必须是
a块的开头，在任何非声明代码之前。
问候，
Russell Hanneken
rg********@pobox.com
删除来自我地址的''g'给我发邮件。
Oops; both i and pt need to be declared at the beginning of main. Unless
you''re writing code to conform to the new (1999) C standard (which you
probably aren''t), local variable declarations have to be at the beginning of
a block, before any non-declaration code.
Regards,
Russell Hanneken
rg********@pobox.com
Remove the ''g'' from my address to send me mail.
" Russell Hanneken" < RG ******** @ pobox.com> schrieb im Newsbeitrag
新闻：a4 ***************** @ newsread4.news.pas.earthl ink.net ...
"Russell Hanneken" <rg********@pobox.com> schrieb im Newsbeitrag
news:a4*****************@newsread4.news.pas.earthl ink.net...
Roman Hartmann写道：
Roman Hartmann wrote:
我想知道是否有更好的方法来访问struct成员
（要点）。
使用p1->访问struct成员; p2-> x看起来相当丑陋且不安全。
I would like to know if there is a better way to access struct members
(the points).
Accessing struct members with p1->p2->x seems rather ugly and unsafe.
[。 。 。]
[. . .]
/ * struct with points，struct profil * /
struct point
{
double x;
double y;
};
struct profil
{struct struct * pt;
struct profil * previous;
struct profil *下一个;
};
int main（无效）
{
int n;
struct profil * p1;
struct profil * begin;
/* struct with points, used by struct profil */
struct point
{
double x;
double y;
};
struct profil
{
struct point *pt;
struct profil *previous;
struct profil *next;
};
int main(void)
{
int n;
struct profil *p1;
struct profil *begin;
[。 。 。]
[. . .]
/ *将一些信息放入结构中* /
p1-> pt-> x = 1265.75;
p1- > pt-> y = 150.73;
p1-> pt ++; / *丑陋，我知道。有一个更好的方法吗？ * /
/* put some information into the struct */
p1->pt->x=1265.75;
p1->pt->y=150.73;
p1->pt++; /* ugly, I know. Is there a better way to do this? */
是的，这个怎么样：
struct point * pt = p1-> pt;
pt [0] .x = 1265.75;
pt [0] .y = 150.73;
Yes, how about this:
struct point *pt = p1->pt;
pt[0].x = 1265.75;
pt[0].y = 150.73;
是的！
处理起来要容易得多现在这样点。
Yes!
It''s much easier to deal with the points now that way.
p1-> pt-> x = 550.55;
p1-> pt-> y = 350.33;
p1-> pt - ;
p1->pt->x=550.55;
p1->pt->y=350.33;
p1->pt--;
pt [1] .x = 550.55;
pt [1]。 y = 350.33;
pt[1].x = 550.55;
pt[1].y = 350.33;
/ *对结构做一些事情* /
/ *而不是打印结果，我打印结构* /
printf（\ nStrukt 1：％f \ n，p1-> pt-> x）;
printf（" \ nStrukt 1：％f \ n"，p1-> pt-> y）;
p1-> pt ++;
printf（" \ nStrukt 2：％f \ n"，p1-> pt-> x）;
printf（" \ n Strukt 2：％f \ n"，p1-> pt-> y）;
返回0;
}
/* do something with the structs */
/* instead of printing results, I print the structs out */
printf("\nStrukt 1: %f\n", p1->pt->x);
printf("\nStrukt 1: %f\n", p1->pt->y);
p1->pt++;
printf("\nStrukt 2: %f\n", p1->pt->x);
printf("\nStrukt 2: %f\n", p1->pt->y);
return 0;
}
int i;
for（i = 0;我< N; ++ i）
{/> printf（" \ nStrukt％d：％f \ n"，i + 1，pt [i] .x）;
printf（" \ nStrukt％d：％f \ n"，i + 1，pt [i] .y）;
}
问候，
Russell Hanneken
int i;
for (i = 0; i < n; ++i)
{
printf("\nStrukt %d: %f\n", i + 1, pt[i].x);
printf("\nStrukt %d: %f\n", i + 1, pt[i].y);
}
Regards,
Russell Hanneken
非常感谢！
Roman
thanks a lot!
Roman
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