一个小而奇怪的<Segmentation Fault>

Question

I am not very experienced with C memory concepts. I tried searching for a solution, but could not find one.

I am just trying to create dynamic array in C. I tried to create it in this way while trying to check whether the addresses are contiguous. The program ran fine.

However, I got a segmentation fault after the statement system("pause") . I also tried to debug using the debugger, with no luck! I'm using Dev CPP.

Can anybody guide me?

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

main()
{
 int a[0], *ptr, i;

 printf("%d", sizeof(a[0]));
 a[0]=1;
 for(i=1;i<10;i++)
 {
  ptr=(int *) malloc(sizeof(int));
  printf("Enter a[%d]:  ", i);
  a[i]= *ptr;
  scanf("%d", &a[i]);

 }
 i=0;
 while(i<10)
 {printf("\n%u", &a[i++]);}
 free(ptr);
 system("pause");
}

Answer 1

int a[0]

doesn't have any space allocated to it (its 0 width)

Yet you write to it on this line:

a[0]=1;

You also assume it has 10 elements here:

while(i<10)
{printf("\n%u", &a[i++]);}
free(ptr);

Practically speaking, as this is just stack memory, you're just writing to another piece of the stack. You could, for example, be overwriting the value of ptr . Often this can go undetected until the stack is unwound and part of it is apparently corruptted. Here the stack is unwound right after system("pause") , when main returns.

(Also, if ptr is overwritten by your writes to a you can't be sure that free does anything reasonable.)

To allocate a 10 integer array, use the syntax:

int a[10];

Then you can use a[0] up to a[9]. But this is C don't expect anything to protect you when you try to read/write to a[10].

Answer 2

There are lots of problems in your code.

1. int a[0]. You are defining an array of 0 elements. In C this will not work. You cannot do a[0]=1 since it will work only if a is an array of 1 or more elements.
2. It gets worse with a[i]=*ptr. 2 problems here, as pointed above a[1], a[2], etc don't exist. and the second problem is you allocated ptr but assigning *ptr which might be having some garbage.

It is just the beginning.

Answer 3

int a[0], ...;
...
a[0]=1;

You are writing out of bounds. This is undefined behavior.

Answer 4

First your a[0] should be a[10] for the way your are trying to use it. Second, just malloc to the pointer and it is basically an array. Though you don't need to do it in a for loop. You can just say ptr = malloc(10 * sizeof(int)); and you will be good. No need to cast it. You can access it's elements through the array subscript like ptr[9] .

Answer 5

Well, your dealing with a is epical:

int a[0];  // no-items array
...

a[0]=1;  // WHAT?
for(i=1;i<10;i++)
{
  ptr=(int *) malloc(sizeof(int));
  a[i]= *ptr; // a was int or pointer? And A[9]?
}

Answer 6

Several problems: 1) Your expecting your array to be 10 elements long but you're declaring it with 0, and make it an array of pointers;

int *a[10];

then

a[i] = (int*)malloc(sizeof(int));

and when freeing;

free(a[i]);