Problems with malloc

Question

int main(int argc, char ** argv)
{
 char *p[10];
 
 for(int i =0 ; i < 10; ++i)
 {
 ssize_t msize = pow(2,i) * sizeof(char);
 p[i] = (char*)malloc(msize);
 printf("%p\n",sbrk(0));
 }
 printf("malloc done!\n");
 
 for(int i =0 ;i < 10; ++i)
 {
 free(p[i]);
 }
 exit(0);
}

The output like that:

0x55ffd4dd5000
0x55ffd4dd5000
0x55ffd4dd5000
0x55ffd4dd5000
0x55ffd4dd5000
0x55ffd4dd5000
0x55ffd4dd5000
0x55ffd4dd5000
0x55ffd4dd5000
0x55ffd4dd5000
malloc done!

When using malloc() to allocate the blocks of memory smaller than MMAP_THRESHOLD bytes, the glibc malloc() will use sbrk() or brk() to finish that. As known, both brk() and sbrk() will adjust the size of the heap.

So if I use malloc() to allocate the small(size < MMAP_THRESHOLD) memory block, the heap should be changed, But why does heap not change in this case?

• Andreas Wenzel: If you want to add information to the question, then please edit the question. Please don't use comments for this, as comments should only be used to respond to other people's comments.
• G.M.: Please edit your question to explain why you think successive calls to sbrk(0) should return different values.
• kaylum: "Why does "the program break" not change". malloc implementations typically have a cache of blocks. It does not need to increase the heap for every malloc call.
• Chris Dodd: most malloc implementations these days use mmap to get memory from the system, rather than brk/sbrk