Memory allocated without allocation using malloc, how?

Question

The following is a simple snippet of code:

int main()
{
int *p;
p=(int*)malloc(sizeof(int));//allocate m/y 4 1 int 
printf("P=%p\tQ=%p",p,p+2);
}

In one run run, he gave me the result, as shown below:

P=0x8210008 Q=0x8210010

The starting address is P is-P = 0x8210008, the next byte is 0x8210009, the next byte is 0x821000A, the next byte is 0x821000B. So 4 bytes for int end. We did not allocate more memory using malloc. Whereas p + 2 leads us to 0x8210010, which is 8 bytes after P (0x8210008).

Answer 1

Firstly, the fact that you typed the address does not mean that memory is allocated at this address. You simply added numbers and produced other numbers.

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Answer 2

. . p+2, , &p[2]. , , char*:

char *highWordAddr = (char*)p + 2;

Answer 3

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Answer 4

. http://www.learncpp.com/cpp-tutorial/68-pointers-arrays-and-pointer-arithmetic/