如何在运行时给c中的结构数组成员的大小

Question

I have a structure which looks like -

struct stack{
 int top;
 char string[size][80];
}stackV;

I want to give user an option to allocate size of char string array at runtime. I had used scanf function to do this -

I tried to achive it by doing -

int size=0;
 struct stack{
  int top;
  char string[size][80];
 } stackV;

But by doing this i got warning which says - variably modified 'string' at file scope

Is there any way by which we can assign size to a structure member array . I can't create a structure inside any function because the structure member is used by other functions also.

Answer 1

You can use a flexible array member to achieve what you need:

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

typedef struct 
{
   int top;
   char string[][80];
} stackv;

int main() 
{
  size_t strings = 3;

  stackv* s = malloc(sizeof(stackv) + strings*80);
  strcpy(s->string[0], "test");
  strcpy(s->string[1], "hello");
  strcpy(s->string[2], "world");

  for(size_t i=0; i<strings; i++)
  {
    puts(s->string[i]);
  }

  free(s);
  return 0;
}

The declaration of char string[][80]; tells the compiler that there will be an unknown amount of char [80] arrays at the end of the struct.

The sizeof(stackv) will therefore only give the size of all other members except the last one (in this case an integer).

Answer 2

You don't know how many elements will be in your stack. Only the user gives you the "size" in the runtime; so you can't allocate an array statically (compilation time), but allocate it dynamically (run time) when getting the size from the user. You are getting an "warning" because you define "size" from size "0" then you try to declare an array of this size and according to the spec there is no array of size 0 as I understand):

ISO 9899:2011 6.7.6.2:

    If the expression is a constant expression, it shall have a value greater than zero.

How to allocate the array dynamically as a struct which contains "size" elements and each element is a string of 80 chars?

#define STR_LEN 80
struct stack {
   int top;
   char *string;
} stackv;

enum {
    SUCCESS = 0,
    FAILURE,
};

int InitStack(stackv *stackv, int size) {
    if (size <= 0)
        return FAILURE;

    stackv->string = (char *)malloc(size*sizeof(char)*80);
    if (stackv->string == NULL)
        return FAILURE;

    stackv->top = 0;

    return SUCCESS;
}

In the main function, you declare the variable "stackv" and pass it by reference (pointer) to the initialization function which I wrote.

Answer 3

This code compiles correctly for me:

int size = 10;

typedef struct 
{
   int top;
   char string[size][80];
} stackv;

I'm using:

$ gcc --version
gcc (Debian 5.4.1-3) 5.4.1 20161019
Copyright (C) 2015 Free Software Foundation, Inc.