[santechz]

Hi there.
I know that malloc() takes its memory chunks from the heap. assuming that there are two consecutive malloc calls, one immedietly after the other, one would assume that the chunks would be allocated nearby like blocks in an array,

what i mean is
//assume sizeof(char)=1
char * p1 = (char*)malloc(sizeof(char)); //now let p1 =1460 (address)
char * p2 = (char*)malloc(sizeof(char));

now p2 should be =1461 (or atleast 1462 for even address boundary), right? but instead, my Turbo Compiler gives the allocated address as 1468 (i.e. an 8 byte gap between the addresses consecutive calls).

experimenting a lot, there always seems to be either a 4 byte or 8 byte gap between the consecutively malloc()'d allocated addresses. , so if we allocated a 3 int array using malloc(3*sizeof(int)), and then one more malloc, the first malloc ()returned address would be 1460, and the second malloc will return 1468.

I searched in the net and found out that ANSI doesnt specify how the memory must be allocated between malloc calls, it is specific to single malloc calls.

could anyone tell me why there is a gap, ?? I do hope u got my doubt....

regards

[C.Ingram]

Try declaring BOTH pointers, then calling malloc twice. My first thought is that the gap between is the pointer p2.

[santechz]

this is how i visualize my question


char * p1 = (char*)malloc(sizeof(char)); //sizeof(char)=1 byte
char * p2 = (char*)malloc(sizeof(char));

now i get p1 = 1460
then p2 must be 1461 (or 1462 for adjustment) if there were no allocation to heap in between these two statements, bcoz this would be the easiest to implement the heap management....

but i get p2 = 1468. in other words, there are 7 bytes which I believe I can use. Why did malloc return 1468 and not 1461(/1462)? Can i really use these 7 bytes. will the heaper (?) allocate these 7 bytes (even when i havent freed p1 and p2) when i request a ..say malloc(4 * sizeof(char))?

regards

[C.Ingram]

Try keeping the code you have, then returning the address of p2 after (not the address of the memory p2 points to). That will anwser a lot of questions.

[fitzjj]

If a char is 1byte, this is 8 bits.

1468 - 1460 = 8

It would make sense if the numbers were bits rather than bytes, since diferent data types use a different number of bits

[santechz]

Hey!
malloc() returns the address of a requested byte(s). you can only request for bytes. Please do try the snippet and see it for your self:
char *ptr1 = (char *) malloc(6*sizeof(char)); //assume 1char=1byte
char *ptr2=(char*) malloc(sizeof(char));//

now there will be either 8 byte difference between ptr1 (not &(ptr1)) and ptr2. i.e. ptr2-ptr1=8. and the "gap" between the 1'st requested(6 bytes) and the successive request will be 2 bytes....

[amit jain]

nice question .. I will work on it