printf参数类型争吵 [英] printf parameter type wrangling

问题描述

我正在终端中打印结构成员表格，我希望

使用带有任意指针的printf，这样代码就更小了。

I不需要切换可能的类型。


但问题是 - 我可以传递任意类型（短，双，字符*，

等）printf作为char *（如果指定的格式字符串是

当然合适）？


我选择''char *''因为我推断它（一个指针）足够大

来容纳作为函数参数传递的任何原始类型。


#include< stdlib.h>

#include< stddef.h>

#include< stdio.h>


void

printf_member（void * obj，size_t off，const char * fmt）

{

char ** p;


p =（char **）（（char *）obj + off）;


printf（fmt，* p）;

}


struct foo {

char c;

浮动f;

int i;

char * s;

};


int

main（无效）

{

struct foo f = {''c''，2.34 f，100，hello，world };


printf_member（& f，offsetof（struct foo，c），" [c] \ n"）;

printf_member（& ; f，offsetof（struct foo，f），" [％f] \ n"）;

printf_member（& f，offsetof（struct foo，i），" [％d ] \ n"）;

printf_member（& f，offsetof（struct foo，s），" [％s] \ n"）;


返回0;

}


不幸的是，这种技术似乎不能作为浮动成员使用

未打印正常。为什么？


$ gcc -Wall -W -ansi -pedantic -o pa pa.c

$ ./pa

[c]

[0.000000]

[100]

[你好，世界]


谢谢，

迈克

I''m printing tables of structure members in a terminal and I would like
to use printf with an arbitrary pointer so that the code is smaller and
I don''t need to switch over ever possible type.

But the question is - can I pass an arbitrary type (short, double, char *,
etc) to printf cast as a char * (provided the specified format string is
appropriate of course)?

I chose ''char *'' because I reasoned it (a pointer) would large enough
to accommodate any primitive type passed as a function parameter.

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

void
printf_member(void *obj, size_t off, const char *fmt)
{
char **p;

p = (char **)((char *)obj + off);

printf(fmt, *p);
}

struct foo {
char c;
float f;
int i;
char *s;
};

int
main(void)
{
struct foo f = { ''c'', 2.34f, 100, "hello, world" };

printf_member(&f, offsetof(struct foo, c), "[c]\n");
printf_member(&f, offsetof(struct foo, f), "[%f]\n");
printf_member(&f, offsetof(struct foo, i), "[%d]\n");
printf_member(&f, offsetof(struct foo, s), "[%s]\n");

return 0;
}

Unfortunately this technique does not appear to work as the float member
was not printed properly. Why?

$ gcc -Wall -W -ansi -pedantic -o pa pa.c
$ ./pa
[c]
[0.000000]
[100]
[hello, world]

Thanks,
Mike

推荐答案

gcc -Wall -W -ansi -pedantic -o pa pa。 c

gcc -Wall -W -ansi -pedantic -o pa pa.c

./ pa

[c]

[0.000000]

[ 100]

[你好，世界]


谢谢，

迈克

./pa
[c]
[0.000000]
[100]
[hello, world]

Thanks,
Mike

Michael B Allen在21/08/05写道：

Michael B Allen wrote on 21/08/05 :

我正在终端打印结构成员表，我想
使用printf一个任意指针，使代码更小，

I''m printing tables of structure members in a terminal and I would like
to use printf with an arbitrary pointer so that the code is smaller and

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