Question

int main()  
{  
    int i=0;  
    int* p_numbers = ((int*) malloc (5*sizeof(int)));  
    int* temp;  
    temp = p_numbers;  
    for(i=0;i<5;i++)  
    {  
        *temp=i;   
        printf("temp %d p_numbers %d",*temp,*p_numbers);  
        temp++;           
    }
}

Please tell that the pointer assigned to temp i.e temp=p_numbers.

DOES temp not point to the same memory position where the p_numbers is pointing to?

Answer 1

    int* p_numbers = ((int*) malloc (5*sizeof(int)));  

              +---+---+---+---+---+ 
p_numbers --> | x | x | x | x | x |
              +---+---+---+---+---+ 


    int* temp;  
    temp = p_numbers;  

p_numbers --+    +---+---+---+---+---+ 
            +--> | x | x | x | x | x |
temp--------+    +---+---+---+---+---+ 

you need also to free p_numbers since otherwise you get a memory leak.

also please make a habit of not casting the return value from malloc because in some cases this can cause hard to find errors

explanation:

the malloc is defined stdlib.h, if you forget to include that header, the malloc function will by default assumed to return int as that is the way in C for functions that have no prototype. Now if you have something like char*p = (char*)malloc(12); this may cause issues because you effectively casting the returned integer to a char*. By explicitly casting you silence the warnings from the compiler and if you have hardware/OS where sizeof(char*) != sizeof(int) you may get a hard to find error so just write

p_numbers = malloc(5*sizeof(int))

if you are using a C++ compiler, use new/delete instead.