• C

What does free() actually do ?

I create a double dimensioned array of int elements and allocate memory to it : -

int **arr;
int x,y;
x = 4;
y = 5;
arr = malloc(x*(sizeof(*arr)));
for (i=0;i<x;i++)
arr[i] = malloc(y*sizeof(**arr));

Then I enter the values in the array.
And then ... free the array : -

for (i=0;i<x;i++) free(arr[i]);

After freeing i print the array and the i still get the values.
Does free() wipes off the allocated memory ?
Does free() sets the pointer to NULL ?
Or it justs flags the allocated memory in a way so that it can be allocated to some other pointers ?

Kindly help
thanks already.


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pls declare like this

int *nos[10];
int no_of_cols = 5;
nos[0] = (int *)malloc(no_of_cols * sizeof(int));
nos[1] = (int *)malloc(no_of_cols * sizeof(int));
>>>After freeing i print the array and the i still get the values.

yes u can because what free does is that it marks the memory pointed to .be available for further allocation calls...

>>>Does free() wipes off the allocated memory ?

 it doenst wipe of .. just marks it as free for allocation ..

>>>Does free() sets the pointer to NULL ?

NO it doesnt ..

>>>Or it justs flags the allocated memory in a way so that it can be allocated to some other pointers ?

yes you got it here

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No the free does not sets memory to NULL. Infact it just releases the allocated memory to the operating system. So that the released memory can be allocated next for another request. Another thing is that you have not done any explicit typecast. In all compilers it will not compile. Try to do typecast. Your some statements, I think, is not correct

     int x,y;
     x = 4;
     y = 6;
     arr = (char **) malloc(x*sizeof(char *));
     for (i=0;i<x;i++)
          arr[i] = (char *) malloc(y*sizeof(char));
// Do your work here
     for (i=0;i<x;i++) free(arr[i]);

Just compare the statements...
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accept answer by akshayxx.


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