变量声明语法 [英] Variable declaration syntax

问题描述

大家好，


就在圣诞节前夕，Chris Torek给了我一些关于闭包的好建议：

< http://groups.google .co.nz / groups？selm = cqcl3k030vj％40news3.newsguy.com& output = gplain>


它包含一个名为fp的函数指针变量的声明：

void（* fp）（void *）;


上面的语法是什么意思？我只能将它消化为void * fp，

即fp是指向void的指针。


谢谢，

Adam

Hi all,

Just before Christmas Chris Torek gave me some great advice about closures in C:
<http://groups.google.co.nz/groups?selm=cqcl3k030vj%40news3.newsguy.com&output =gplain>

It includes this declaration of a function pointer variable called fp:
void (*fp)(void *);

What does the syntax above mean? I can only digest it as void * fp,
i.e. fp is a pointer to void.

Thanks,
Adam

推荐答案

>就在圣诞节前夕，Chris Torek在C中给出了一些关于闭包的好建议：

> Just before Christmas Chris Torek gave me some great advice about closures in C:

< http://groups.google.co.nz/groups？selm = cqcl3k030vj％40news3.newsguy.com& output = gplain>

它包含了一个名为fp的函数指针变量的声明：
void（* fp）（void *）;

上面的语法是什么意思？我只能将其消化为void * fp，
即fp是指向void的指针。

<http://groups.google.co.nz/groups?selm=cqcl3k030vj%40news3.newsguy.com&output =gplain>

It includes this declaration of a function pointer variable called fp:
void (*fp)(void *);

What does the syntax above mean? I can only digest it as void * fp,
i.e. fp is a pointer to void.

fp是一个指向函数的指针，该函数返回带参数的void清单

由空格*组成。


例如：


void do_naught（void *）;

void do_aught（void *）;


.....


if（bla）

fp = do_aught;

其他

fp = do_naught;


.....


/ *通过fp * /

（* fp）（地址）调用相应的函数;


.... 。


（* fp）（another_one）;


.....


优点：您不必要求bla所有的时间，但设置

fp到你需要的功能。


如果你有C声明的问题，如

int （* p）[10];

vs

int * p [10];

你可以搜索程序cdecl，它什么都不做但

翻译乱码< - >英文，其中乱码是C声明。

干杯

Michael

-

电子邮件：我的是一个gmx点地址。

fp is a pointer to a function returning void with a parameter list
consisting of a void *.

example:

void do_naught (void *);
void do_aught (void *);

.....

if (bla)
fp = do_aught;
else
fp = do_naught;

.....

/* Call the appropriate function through fp */
(*fp)(address);

.....

(*fp)(another_one);

.....

Advantage: You have not to ask for "bla" all the time but set
fp to the function you need.

If you have problems with C declarations such as
int (*p)[10];
vs
int *p[10];
you could search for the program cdecl which does nothing but
translate gibberish<->english where gibberish are C declarations.
Cheers
Michael
--
E-Mail: Mine is a gmx dot de address.

" Adam华纳" <我们**** @ consulting.net.nz>在留言中写道

news：pa ************************** @ consulting.net.n z ...

"Adam Warner" <us****@consulting.net.nz> wrote in message
news:pa**************************@consulting.net.n z...

大家好，

就在圣诞节前，Chris Torek给了我一些关于闭包的好建议
in C：
< http://groups.google.co.nz/groups？ se ... wsguy.com& outp

ut = gplain>
它包含一个名为fp的函数指针变量的声明：
void（* fp） （void *）;

上面的语法是什么意思？我只能将其消化为void * fp，
即fp是指向void的指针。

Hi all,

Just before Christmas Chris Torek gave me some great advice about closures in C: <http://groups.google.co.nz/groups?se...wsguy.com&outp
ut=gplain>
It includes this declaration of a function pointer variable called fp:
void (*fp)(void *);

What does the syntax above mean? I can only digest it as void * fp,
i.e. fp is a pointer to void.

''fp''是指向函数的指针采用类型为

''void *''的单个参数，并且不返回值。在网上搜索实用工具

称为''cdecl''，它可以接受声明并告诉你

的含义。

- Mike

''fp'' is a pointer to a function which takes a single argument of type
''void *'', and does not return a value. Search the web for a utility
called ''cdecl'', which can take a declaration and tell you what
it means.
-Mike

嗨Michael Mair，

On Sun，2005年1月2日13:10:56 +0100，Michael Mair写道：

Hi Michael Mair,
On Sun, 02 Jan 2005 13:10:56 +0100, Michael Mair wrote:

就在圣诞节前夕，Chris Torek给了我一些关于闭包的好建议：
< http://groups.google.co.nz/groups？selm = cqcl3k030vj ％40news3.newsguy.com& output = gplain>

它包含一个名为fp的函数指针变量的声明：
void（* fp）（void *）;

Just before Christmas Chris Torek gave me some great advice about closures in C:
<http://groups.google.co.nz/groups?selm=cqcl3k030vj%40news3.newsguy.com&output =gplain>

It includes this declaration of a function pointer variable called fp:
void (*fp)(void *);

fp是一个函数返回void的指针，参数列表由void *组成
。

fp is a pointer to a function returning void with a parameter list
consisting of a void *.

谢谢，我终于弄清楚了语法！如下所示：


#include< stdio.h>

#include< stdlib.h>

#include < time.h>


unsigned int uinc1（int x）{return ++ x; }


unsigned int uinc2（int x）{return x + = 2; }


int main（）{

//种子随机数生成器，便携式

struct tm epoch;

epoch.tm_year = 27; epoch.tm_mon = 9-1; epoch.tm_mday = 4; epoch.tm_isdst = -1;

epoch.tm_hour = 0; epoch.tm_min = 0; epoch.tm_sec = 0;

clock_t current_clock = clock（）;

time_t current_time = time（& current_clock）;

time_t time_since_epoch = mktime（& epoch）;

double diff = difftime（current_time，time_since_epoch）;

unsigned int seed =（unsigned int）diff;

printf（种子是％u \ n，种子）;

srand（种子）;


unsigned int（* fp）（int ）;

//注意：伪随机数生成器可能很差

if（rand（）％2 == 0）fp = uinc1;否则fp = uinc2;

unsigned int sum = fp（1）;

printf（" Sum is％u \ n"，sum）;


返回0;

}


BTW我发现cdecl是为Debian打包的。


问候，

Adam

Thanks, I''ve finally grokked the syntax! As demonstrated below:

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

unsigned int uinc1 (int x) { return ++x; }

unsigned int uinc2 (int x) { return x+=2; }

int main() {
//seed random number generator, portably
struct tm epoch;
epoch.tm_year=27; epoch.tm_mon=9-1; epoch.tm_mday=4; epoch.tm_isdst=-1;
epoch.tm_hour=0; epoch.tm_min=0; epoch.tm_sec=0;
clock_t current_clock=clock();
time_t current_time=time(&current_clock);
time_t time_since_epoch=mktime(&epoch);
double diff=difftime(current_time, time_since_epoch);
unsigned int seed=(unsigned int) diff;
printf("Seed is %u\n", seed);
srand(seed);

unsigned int (* fp) (int);
//Note: The pseudo random number generator may be poor
if (rand()%2==0) fp=uinc1; else fp=uinc2;
unsigned int sum=fp(1);
printf("Sum is %u\n", sum);

return 0;
}

BTW I found cdecl was packaged for Debian.

Regards,
Adam

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