按编号引用结构成员。 [英] referring to struct members by number.
问题描述
问候
对于某些操作，我希望能够轻松访问struct
成员。这是一个例子。
struct mystruct {
char member1 [3];
char member1 [50];
char member1 [12];
};
/ *打印结构中的字符串* /
for（i = 0; i< 3; i ++）{
switch（i）：
case 0：
printf（"％s\r\ n"，mystruct.member1）;
休息;
案例0：
printf （％s\r\ n，mystruct.member2）;
休息;
案例0：
printf（" ;％s \\ n"，mystruct.member3）;
休息;
}
这是非常cumbersom，特别是大型结构。是否有某种方式
我可以避免为每个成员命名？
类似下面的伪代码;
for（i = 0; i< 3; i ++）{
打印会员我
}
我很感激这方面的建议。
问候。
edson写道：
< blockquote class =post_quotes>
问候
对于某些操作，我希望能够轻松访问struct
成员。这是一个例子。
struct mystruct {
char member1 [3];
char member1 [50];
char member1 [12];
};
/ *打印结构中的字符串* /
for（i = 0; i< 3; i ++）{
switch（i）：
case 0：
printf（"％s\r\ n"，mystruct.member1）;
休息;
案例0：
printf （％s\r\ n，mystruct.member2）;
休息;
案例0：
printf（" ;％s \\ n"，mystruct.member3）;
休息;
}
这是非常cumbersom，特别是大型结构。是否有某种方式
我可以避免为每个成员命名？
类似下面的伪代码;
for（i = 0; i< 3; i ++）{
打印会员我
}
我很感激这方面的建议。
一个ptr到char的数组？
char * jagged_array [3];
jagged_array [0] = malloc（3）;
if（！jagged_array [0]）
abort（）;
jagged_array [1] = malloc（50）;
if（！jagged_array [1]）
abort（）;
jagged_array [2] = malloc（12）;
if（！jagged_array [2]）
abort（）;
显然这可以变成一个函数
打印： -
for（i = 0; i< 3; i ++）
printf（"％s" jagged_array [i]）;
代码未编译
-
Nick Keighley
Nick Keighley< ni ****************** @hotmail.comaécrit*：
>
edson写道：
>问候
对于某些操作，我希望能够轻松访问struct
成员。这是一个例子。
struct mystruct {
char member1 [3];
char member1 [50];
char member1 [12];
对于（i = 0; i< 3; i ++）{
switch（i）：打印struct * /
中的字符串;}
案例0：
printf（"％s\r\ n"，mystruct.member1）;
休息;
案例0：
printf（" ;％s \\\\ n"，mystruct.member2）;
中断;
案例0：
printf（"％s\r\ n"，mystruct.member3 ）;
休息;
}
这是非常麻烦的，尤其适用于大型结构。是否有某种方式
我可以避免为每个成员命名？
类似下面的伪代码;
for（i = 0; i< 3; i ++）{
print member i
}
一个ptr到char的数组？
3动态分配，3次失败测试和3次malloc开销:-(
我真的更喜欢
struct mystruct {
char member1 [3];
char member1 [50];
char member1 [12];
char * memtab [3];
};
void init_mystruct（struct mystruct * s）{
s - > memtab [0] =&（s-> member1）;
s-> memtab [1] =&（s-> member2）;
s-> memtab [2] =&（s-> member3）;
}
Marc Boyer
我使用的解决方案是（手）编写一个指针数组。这个
的缺点是如果更改了结构，必须手动编辑它。
是否存在另一个编译器可能会将成员置于不同于结构起点的不同偏移量的位置的风险？我是
使用GCC，我发现成员完全连续，没有
差距恩他们。
struct mystruct {
char member1 [3];
char member2 [50];
char member3 [12];
};
s_l_fieldpos [] = {0,3,50};
/ *打印结构中的字符串* /
for（i = 0; I< 3; i ++）{
printf（"％s\r\ n"，mystruct.member1 [s_l_fieldpos [i]]）;
}
Greetings
For certain operations I would like to have easy access to struct
members. Here is an example.
struct mystruct {
char member1[3];
char member1[50];
char member1[12];
};
/* print the strings in the struct */
for(i=0; i<3; i++){
switch(i):
case 0:
printf(" %s\r\n", mystruct.member1 );
break;
case 0:
printf(" %s\r\n", mystruct.member2 );
break;
case 0:
printf(" %s\r\n", mystruct.member3 );
break;
}
This is very cumbersom, sepecially for large structs. Is there some way
I could avoid having to name each member?
something like the following pseudocode;
for(i=0; i<3; i++){
print member i
}
I would appreciate advice on this.
Regards.
edson wrote:Greetings
For certain operations I would like to have easy access to struct
members. Here is an example.
struct mystruct {
char member1[3];
char member1[50];
char member1[12];
};
/* print the strings in the struct */
for(i=0; i<3; i++){
switch(i):
case 0:
printf(" %s\r\n", mystruct.member1 );
break;
case 0:
printf(" %s\r\n", mystruct.member2 );
break;
case 0:
printf(" %s\r\n", mystruct.member3 );
break;
}
This is very cumbersom, sepecially for large structs. Is there some way
I could avoid having to name each member?
something like the following pseudocode;
for(i=0; i<3; i++){
print member i
}
I would appreciate advice on this.an array of ptr to char?
char* jagged_array [3];
jagged_array[0] = malloc(3);
if (!jagged_array[0])
abort();
jagged_array[1] = malloc(50);
if (!jagged_array[1])
abort() ;
jagged_array[2] = malloc(12);
if (!jagged_array[2])
abort();
obviously this can be turned into a function
printing:-
for (i = 0; i < 3; i++)
printf ("%s" jagged_array[i]);
code not compiled
--
Nick Keighley
Nick Keighley <ni******************@hotmail.coma écrit*:>
edson wrote:>Greetings
For certain operations I would like to have easy access to struct
members. Here is an example.
struct mystruct {
char member1[3];
char member1[50];
char member1[12];
};
/* print the strings in the struct */
for(i=0; i<3; i++){
switch(i):
case 0:
printf(" %s\r\n", mystruct.member1 );
break;
case 0:
printf(" %s\r\n", mystruct.member2 );
break;
case 0:
printf(" %s\r\n", mystruct.member3 );
break;
}
This is very cumbersom, sepecially for large structs. Is there some way
I could avoid having to name each member?
something like the following pseudocode;
for(i=0; i<3; i++){
print member i
}an array of ptr to char?
3 dynamic allocations, 3 failure tests and 3 malloc overhead :-(
I would really prefer
struct mystruct {
char member1[3];
char member1[50];
char member1[12];
char* memtab[3];
};
void init_mystruct( struct mystruct* s){
s->memtab[0]= &(s->member1);
s->memtab[1]= &(s->member2);
s->memtab[2]= &(s->member3);
}
Marc Boyer
The solution I am using is to (hand) write an array of pointers. This
has the disadvantage that it must be hand edited if the struct is changed.
Is there a risk that another compiler might put the members in different
locations with different offsets from the start of the struct? I am
using GCC and I find that the members are perfectly contiguous with no
gaps between them.
struct mystruct {
char member1[3];
char member2[50];
char member3[12];
};
s_l_fieldpos[] = {0, 3, 50};
/* print the strings in the struct */
for(i=0; i<3; i++){
printf(" %s\r\n", mystruct.member1[s_l_fieldpos[i]]);
}
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