Can realloc shrink my array on the left side (C only)?

Question

I want to move a large chunk of data i've got in my memory. Unfortunately this data is saved as an array, and i cannot change that. I can't use circular arrays, because the same memory is also used by a couple of fortran methods i do not want to change. On top of it, the arrays are accessed very very often in between the movement. So i can do this:

int *array = (int*) malloc(sizeof(int)*5);
int *array2=NULL;
//Now i want to move my data one step to the left
array=(int*) realloc(array,6);
array2=array+1;
memmove(array,array2,5*sizeof(int));
array=(int*) realloc(array,5);

This should work fine, but it looks so wasteful ;). If i could tell my compiler to take away data on the left side of a shrinking array, my data would sort of creep through the memory, but i wouldn't have to do any copying. Like this:

int *array = (int*) malloc(sizeof(int)*5);
//Now i want to move my data one step to the left
array=(int*) realloc(array,6);
array=(int*) realloc_using_right_part_of_the_array(array,5);

So basically i want to finish with a pointer to array+1, with the 4 byte left of it freed. I played around with free() and malloc() but it didn't work... I'm aware that the realloc could also result in a memcpy call, but not everytime! So it could be faster, couldn't it?

Answer 1

No. There is no way to give back the lower part of the memory you allocated. Also, your original code is wrong, since you're copying indeterminate memory.

int *array = (int*) malloc(sizeof(int)*5);
// Fill memory:
// array - {'J', 'o', h', 'n', '\0'}; 
int *array2=NULL;
//Now i want to move my data one step to the left
array=(int*) realloc(array,6);
// array - {'J', 'o', h', 'n', '\0', X};
array2=array+1;
// array2 pointer to 'o of array.
memmove(array,array2,5*sizeof(int));
// This copies the indeterminate x:
// array - {'o', h', 'n', '\0', X, X}
array=(int*) realloc(array,5);
// array - {'o', h', 'n', '\0', X}

X means indeterminate.