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在同一结构中初始化指向数组的常量指针

[英]Initialize constant pointer to array in same struct

typedef struct {
    char a[100];
    char const *secondHalfOfA;
}
//Can i set { secondHalfOfA = a[49]; } somehow?

Sure you can. 你当然可以。

typedef struct {
    char a[100];
    char const *const secondHalfOfA;
} toto;
#define TOTO_INITIALIZER(NAME) { .secondHalfOfA = &((NAME).a[49]) }

static toto A = TOTO_INITIALIZER(A);

Using initializers like that consistently has the extra plus that it initializes your array with all 0 , too. 始终如一地使用这样的初始化器具有额外的优点,那就是它也用全0初始化了数组。

This is for C99 with designated initializers. 这适用于具有指定初始化程序的C99。 If you only have historical C you could do something like 如果您只有历史C,则可以执行以下操作

#define TOTO_INITIALIZER(NAME) { { 0 }, &((NAME).a[49]) }

Edit: Seeing your comment on another answer, I suppose that you had the const in your type on the wrong side. 编辑:看到您对另一个答案的评论,我想您输入错误的类型就是const But real C initialization works very well for that case, too. 但是在这种情况下,真正的C初始化也非常有效。

You cannot automatically initialize a pointer to the middle of an object. 您不能自动初始化指向对象中间的指针。

You can, however, pretend to have overlapping objects. 但是,您可以假装有重叠的对象。 It's not pretty though :-) 虽然不是很漂亮:-)

#include <stdio.h>
#include <string.h>

struct twoparts {
  union {
    char a[20];
    struct {
      char dummy[10];
      char b[10];
    } s;
  } u;
};

int main(void) {
  struct twoparts x;
  strcpy(x.u.a, "1234567890123456789");
  printf("second part: %s\n", x.u.s.b);
  return 0;
}

No. You can't do that in a typedef . 不,您不能在typedef做到这一点。 Remember: with typedef keyword you are defining data types, not variables. 请记住:使用typedef关键字定义的是数据类型,而不是变量。

In C++ you can do that in the default constructor. 在C ++中,您可以在默认构造函数中执行此操作。 But in C, there is no way to do that. 但是在C语言中,没有办法做到这一点。

Is Pablo said , you can't do that with an initializer. Pablo说过 ,您不能使用初始化程序来做到这一点。 You can do it easily enough after initialization, but it has to be every time: 初始化后,您可以很容易地做到这一点,但是每次都必须这样做:

// Definition (once)
typedef struct {
    char a[100];
    char const *secondHalfOfA;
} TYPENAME;

// Use (each time)
TYPENAME t;
t.secondHalfOfA = &t.a[49];

// Or
TYPENAME *pt;
pt = malloc(sizeof(*pt));
pt->secondHalfOfA = &pt->a[49];

It's for reasons like this that we have object oriented languages like C++ (and Java and ...) and their associated constructors , so that we can create structures with custom initializations reliably. 出于这样的原因,我们拥有面向对象的语言(例如C ++(以及Java和...)及其关联的构造函数) ,以便我们可以可靠地创建具有自定义初始化的结构。


Demonstration program: 示范程序:

#include <stdio.h>

// Definition (once)
typedef struct {
    char a[100];
    char const *secondHalfOfA;
} TYPENAME;

int main(int argc, char* argv[])
{
    // Use of the type (need to initialize it each time)
    TYPENAME t;
    t.secondHalfOfA = &t.a[49];

    // Example actual use of the array
    t.a[49] = 'A';
    printf("%c\n", *t.secondHalfOfA); // Prints A

    // Uncommenting the below causes a compiler error:
    // "error: assignment of read-only location ‘*t.secondHalfOfA’"
    // (That's gcc's wording; your compiler will say something similar)
    //*t.secondHalfOfA = 'B';
    //printf("%c\n", *t.secondHalfOfA);

    return 0;
}

Compilation and output using gcc: 使用gcc编译和输出:

$ gcc -Wall temp.c
$ ./a.out
A

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