Question

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

int main()
{ 
  char* charPtr="see me";
  printf("%s\n", charPtr);
  printf("%d", charPtr);
  return 0;
}

I do not understand the line:
char* charPtr=”see me”;

how is the memory allocated?

Is 7 bytes of memory allocated and the pointer is allocated to the first byte? and

I know the sizeof(char*) is 4bytes and how does it matter here and what does it influence in allocation of memory?

Some one please help me.thanks…

Answer 1

charPtr is a pointer to an string literal “see me” which resides somewhere in implementation defined memory region. This string literal should not be modified and any attempt to do so leaves you with an Undefined Behavior.

I know the sizeof(char*) is 4 bytes and how does it matter here and what does it influence in allocation of memory?

Since charPtr is a pointer you cannot use sizeof() to determine the length of the string literal. If you do so, What you get is the memory occupied by the pointer and not by the string. You will need to use strlen() if you need to get the length of the string.

         +-----+     +---+---+---+---+---+---+----
charPtr: |  *======> | s | e | e |   | m | e |\0 |
         +-----+     +---+---+---+---+---+---+----