What does pointer array from malloc mean?

Question

unsigned char *dup = malloc(size);

My question may be naive. What does dup[2] mean? Is it a pointer to the third char from the malloced memory or it's the value of the third char from the malloced memory? I have searched google but found no result to explain this. Many thanks for your time.

Answer 1

dup[2] is sematically identical to *(dup + 2) . So it is the value of the third byte pointed to by dup . That is, the memory addresses are:

dup, dup+1, dup+2, ....., dup+size-1

Note that malloc does not initialize the returned memory, so strictly speaking, the value of dup[2] could be anything.

Answer 2

it's the value of the third char from the malloced memory?

This.

dup[2] equivalent to *(dup + 2) . The + 2 implicitly acts like + 2 * sizeof(char) .

If you want a pointer to the third char in the memory, without dereferencing it, then you just use the same as above. without the dereferencing operator:

unsigned char *thirdChar = dup + 2;

Answer 3

Clearly, dup[k] in this case representing the third character of the string, which is very much similiar to *(dup + 2) .

The supporting code as follows:

#include<stdio.h>
#include<string.h>

int main() {

unsigned char *dup = malloc(10);
scanf("%s", dup);
printf("%c", dup[2]);
printf("\n%c", *(dup+2));
   return 0;
}

The output being the same for both printf statement, it made it very clear.