C Memset输出

Question

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

int main() {
  char* p = new char[10];
  memset(p,0,10);
  printf("%c",*p);
}

I suppose memset set every byte starting p to 0 . I'm a little surprised to see nothing at all printed out. What on earth was happening for memset ?

Answer 1

memset does set all the bytes to 0; thus, when you dereference p , you get a char with value 0 (the NUL byte), and on most systems, printing such a char produces no visible output. If you want to print the numeric value of the byte instead, use printf("%d", *p); .

Answer 2

ANS: 0 (int data) is typecasted to equivalent char (ASCII) and then copied on all 10 arrays of memory. Coincidently the equivalent of 0 (int) to char is null character `'\\0'. So, nothing is shown on the screen. Logically we can say that null is printed in screen.

{ie; (char)0 is equivalent to '\0' (null character)}

EXPLAINATION:

memset(p,0,10);

Observer the second parameter 0, which is an integer (of 2 bytes), however the memset() will have to set the 0 data in each byte array of p. How 2 byte integet can be copied on 1 byte memory space? This is not possible.

So, memset() method firstly typecast the int data to char (1 byte) and then writes the char(1 byte) to each byte of that memory array.

Note: memset() will not put any null character at the end of string. So we have to do it explicitly.