在结构中动态分配字符串

Question

If I have a struct defined as such:

typedef struct{

    char a[];

}my_struct_t;

How do you allocate memory for a string with malloc() such that it is stored in my_struct_t ?

Answer 1

Code could uses a flexible array member to store string in the structure. Available since C99. It requires at least one more field in the struct .

As a special case, the last element of a structure with more than one named member may have an incomplete array type; this is called a flexible array member . ... C11 §6.7.2. 18

typedef struct fam {
  size_t sz;  // At least 1 more field needed. 
  char a[];   // flexible array member - must be last field.
} my_struct_t;

#include <stdlib.h>
#include <string.h>

my_struct_t *foo(const char *src) {
  size_t sz = strlen(src) + 1;

  // Allocate enough space for *st and the string
  // `sizeof *st` does not count the flexible array member field 
  struct fam *st = malloc(sizeof *st + sz);
  assert(st);

  st->sz = sz;
  memcpy(st->a, src, sz);
  return st;
}

---

As exactly coded, the below is not valid C syntax. Of course various compilers offer language extensions.

typedef struct{
    char a[];
}my_struct_t;

Answer 2

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

typedef struct {
     char* a;
} my_struct_t;


int main() 
{
    my_struct_t* s = malloc(sizeof(my_struct_t));
    s->a = malloc(100);
    // Rest of Code
    free(s->a);
    free(s);
    return 0;
}

Array of 100 chars.