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Say I declared an array[10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Later I want it to be array[8] = {2, 3, 4, 5, 6, 7, 8, 9}.

Dismiss the first 2 elements. So it would start on array[2]. Reallocing array to array[2].

I tried: 

int *array=(int*)malloc(10*sizeof(int));
...//do stuffs
array=(int*)realloc(array[2],8*sizeof(int));

It didn't work. Neither using &array[2], *array[2], nor creating an auxiliary array, reallocing Array to AuxArr than free(AuxArr).

Can I get a light?

Bjorn Reppen
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2 Answers2

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You can only realloc a pointer to a memory block that has already been alloc'ed. So you can realloc(array), but not array[2] since that is a pointer to a location in the middle of a memory block.

You may want to try memmove instead.

Edit:In response to ThingyWotsit's comment, after memomving the data you want to the front of the array, then you can realloc to drop off the tail end.

Brad Parker
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Just use array += 2 or array = &array[2]. You can't realloc() it.

Gold Dragon
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