c - Is returning a local pointer cause undefined behaviour -

i have doubt in statement

p = my_malloc(4);

my_malloc has local pointer called p, when function returns address of pointer deallocated. how int* p in main hold address returned function. when function returns, address used may or may not used other functions or processes. below program undefined behaviour?

#include<stdio.h> #include<unistd.h>  void* my_malloc(size_t size){  void *p;  p = sbrk(0);   p = sbrk(size); // give previous address  //p = sbrk(0); // give current address  if(p != (void *)-1){    printf("\n address of p : 0x%x \n",(unsigned int)p);  }  else{   printf("\n unable allocate memory! \n");   return null;  }  return p; }  int main(){  int* p;  p = my_malloc(4);  printf("\n address of p : 0x%x \n",(unsigned int)p); } 

your code ok, beware sbrk(2) obsolete (and thread unfriendly), malloc implementations use mmap(2) instead.

what undefined behavior return address of local variable,

void* topofstack() {    char c;    return &c; } 

and recent gcc compilers (e.g. 4.8) make warning, @ least -wall should use. regarding call stacks see this answer gives lot of useful links.

when coding malloc, code free (and try avoid making syscall often, re-use free-d memory in malloc when possible). source code of existing malloc free software implementations. musl libc has quite readable malloc/ ...