malloc和功能 [英] malloc and functions

问题描述

我遇到了malloc和gcc的问题。有什么问题

我的代码还是这个编译器错误？


我正在运行这个程序：


#include< stdio.h>

#include< stdlib.h>


typedef struct pxl {

double lon，lat;

double x，y，z;

double px，py;

}像素;

struct chn {

int index;

struct nde * node;

};


struct nde {

像素位置;

struct chn * forward;

struct chn * reverse;

int * chain_num;

int size;

};


typedef struct chn chain;

typedef struct nde node;


node * createnode（）;


main（）

{


node * startn;


startn = createnode（）;


startn-> size = 2;

startn-> forward = malloc（2 * sizeof（chain *））;

startn-> reverse = malloc（2 * sizeof（chain *））;


printf（" sf％p\ nsr％p\\\
\ n"，

startn-> ;转发，

开始 - >反转

）;


printf（" sf％p \ nsr％p \ n \\ nn，&＃;

startn->转发，

startn->反转

）;
< br $>
}


节点* createnode（）

{

节点*节点;

node = malloc（sizeof（node））;

返回节点;

}


我得到以下内容输出。


sf 0x660168

sr 0x660178

sf 0xa383731

sr 0x660178


startn->前进点在两个printfs之间发生变化，尽管

它们之间没有代码。

解决方案

raphfrk< ra ***** @ netscape.netwrites：

>

node * createnode（）

{

node * node;

node = malloc（sizeof（node））;

只需快速浏览一下，上面的混淆了我。


节点？试试node * pnode ...


不知道是否能解决你的问题。

7月15日，4：09 pm，Richard< rgr ... @ gmail.comwrote：

raphfrk< raph ... @ netscape.netwrites：

node * createnode（）

{

node * node;

node = malloc（sizeof（node） ）;

只需快速浏览一下，上面的混淆了我。


节点？试试node * pnode ...


不知道它是否能解决您的问题。

是的，修正了它。我一直试图弄清楚它多年来

（大约3个小时）。 gcc必须误解我想要的东西。它虽然没有给出警告，但是b $ b没有发出警告。


非常感谢。

< blockquote> 7月15日上午10:57，raphfrk< raph ... @netscape.netwrote：

我遇到了malloc和gcc的问题。有什么问题

我的代码还是这个编译器错误？


我正在运行这个程序：


#include< stdio.h>

#include< stdlib.h>


typedef struct pxl {

double lon，lat;

double x，y，z;

double px，py;

}像素;


struct chn {

int index;

struct nde * node;

};


struct nde {

像素位置;

struct chn * forward;

struct chn * reverse;

int * chain_num;

int size;

};


typedef struct chn chain;

typedef struct nde node;


node * createnode（）;


main（）

''int main（void）''更好。

{


节点* startn;


startn = createnode（）;


startn-> size = 2;

startn-> forward = malloc（2 * sizeof（chain *））;

startn-> reverse = malloc（2 * sizeof（链*） ））;


printf（" sf％p\ nsr％p\\\
\ n"，

startn-> forward，

startn->反转

）;


printf（" sf％p\ nsr％p\ nn \\ n"，

startn-> forward，

startn-> reverse

）;

％p转换说明符需要一个void *参数，你需要

适当地投射你的参数。

}


node * createnode（）

{

node * node;

哎呀。对typedef和变量使用相同的名称（特别是一个

是指向该类型的指针）是丑陋的，并且有可能导致

很大的混乱，请参阅下面。

node = malloc（sizeof（node））;

提示：sizeof是否适用于typedef名称或变量名称？

这可能不是你所期望的。 />
您还应该检查malloc的返回值是否失败。

返回节点;

}


我得到以下输出。


sf 0x660168

sr 0x660178


sf 0xa383731

sr 0x660178

startn->前进点在两个printfs之间发生变化，尽管

之间没有代码他们。

修复上述问题，如果您仍有问题，请告诉我们。


Robert Gamble

I am having an issue with malloc and gcc. Is there something wrong
with my code or is this a compiler bug ?

I am running this program:

#include <stdio.h>
#include <stdlib.h>

typedef struct pxl {
double lon, lat;
double x,y,z;
double px,py;
} pixel;
struct chn {
int index;
struct nde *node;
};

struct nde {
pixel position;
struct chn *forward;
struct chn *reverse;
int *chain_num;
int size;
};

typedef struct chn chain;
typedef struct nde node;

node *createnode( );

main()
{

node *startn;

startn = createnode( );

startn->size = 2;
startn->forward = malloc( 2 * sizeof( chain * ) );
startn->reverse = malloc( 2 * sizeof( chain * ) );

printf("sf %p\nsr %p\n\n",
startn->forward,
startn->reverse
);

printf("sf %p\nsr %p\n\n",
startn->forward,
startn->reverse
);

}

node *createnode( )
{
node *node;
node = malloc( sizeof(node) );
return node;
}

I get the following output.

sf 0x660168
sr 0x660178

sf 0xa383731
sr 0x660178

The startn->forward point changes between the two printfs despite
there being no code between them.

解决方案

raphfrk <ra*****@netscape.netwrites:

>
node *createnode( )
{
node *node;
node = malloc( sizeof(node) );

Just a quick glance, the above "confused me".

node? Try "node * pnode" ...

No idea if it will fix your problem.

On Jul 15, 4:09 pm, Richard <[e-mail]:

raphfrk <[e-mail]:

node *createnode( )
{
node *node;
node = malloc( sizeof(node) );

Just a quick glance, the above "confused me".

node? Try "node * pnode" ...

No idea if it will fix your problem.

Yeah, that fixed it. I have been trying to figure it out for ages
(well about 3 hours). gcc must be misinterpreting what I wanted. It
didn''t give a warning though.

Thanks alot.

On Jul 15, 10:57 am, raphfrk <[e-mail]:

I am having an issue with malloc and gcc. Is there something wrong
with my code or is this a compiler bug ?

I am running this program:

#include <stdio.h>
#include <stdlib.h>

typedef struct pxl {
double lon, lat;
double x,y,z;
double px,py;
} pixel;

struct chn {
int index;
struct nde *node;
};

struct nde {
pixel position;
struct chn *forward;
struct chn *reverse;
int *chain_num;
int size;
};

typedef struct chn chain;
typedef struct nde node;

node *createnode( );

main()

''int main (void)'' is better.

{

node *startn;

startn = createnode( );

startn->size = 2;
startn->forward = malloc( 2 * sizeof( chain * ) );
startn->reverse = malloc( 2 * sizeof( chain * ) );

printf("sf %p\nsr %p\n\n",
startn->forward,
startn->reverse
);

printf("sf %p\nsr %p\n\n",
startn->forward,
startn->reverse
);

The %p conversion specifier expects a void * argument, you need to
cast your arguments appropriately.

}

node *createnode( )
{
node *node;

Yuck. Using the same name for a typedef and a variable (esp. one that
is a pointer to that type) is ugly and has the potential to cause
great confusion, see below.

node = malloc( sizeof(node) );

Hint: Does sizeof apply to the typedef name or the variable name?
It''s probably not what you expect.
You should also check the return value of malloc for failure.

return node;
}

I get the following output.

sf 0x660168
sr 0x660178

sf 0xa383731
sr 0x660178

The startn->forward point changes between the two printfs despite
there being no code between them.

Fix the issues mentioned above and let us know if you still have a
problem.

Robert Gamble

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