c, c++ most basic double quotes

Question

char* a="HELLO WORLD";

IF ADDRESS of 'H' is 0x01 then the printf with %s prints to D but if the same code is written with manual printing routine

while(*a!=NULL) {printf("%c",n[a]);n++;}

this prints a few more characters.. but

printf("%s",a);

prints it perfectly.

while(*a++) printf("%c", *(a-1)); or 

for(;*a++;)printf("%c", *(a-1));

although work but i dont want solutions but the process mechanisms..

so the question coming to my mind is

whether printf gets the length of the string from some register(or any memory unit) or it performs character check.. then prints...

Answer 1

did you mean or you have error:

int main() {
   int n = 0;
   char* a="HELLO WORLD";

   while(a[n] != NULL) {printf("%c",a[n]);n++;}
}

explanation about what is wrong:

while(*a!=NULL) printf("%c",n[a]);n++;

1. a is not modified anywhere, so *a will not change it's value.
2. Although n[a] is perfectly valid construct in C I strongly recommend not to use it, because it is semantically incorrect. You access array by index, not index by array.
3. You increment index (n++) but check the pointer to array. You could possibly increment a inself like this: while(*a!=NULL) {printf("%c",*a);a++;}

Answer 2

The way you're indexing into the character string is odd. It works for the string, but won't stop because you never change the value of *a. What your program does is try to get the a offset of n, so for the first 11 positions they are the same, but the loop doesn't terminate because *a will always be 'H'. What you'd want the terminating condition to be is n < strlen(a).

However, the more succinct way to write that program would be:

int main(int argc, char **argv) {
    char *a = "HELLO WORLD";
    while(*a) printf("%c", *a++);
    return 0;
}

This works because a is an array of characters and as we're printing out each character (de-referencing the value stored at the position) we also increment to the next position. The string should terminate with a NULL reference, which will cause the loop to terminate sine *a == 0 at the NULL terminator.

Answer 3

The corect way to do this is:

#include <iostream>
using namespace std;

int main() {
    char *a="HELLO WORLD";
    int n = 0;
    while(a[n]!=NULL){
        cout<<a[n];
        n++;
    }

   cout<<'\n';
   return 0;
}

Answer 4

As far as I remember, by default when you created char* a it will be something as {HELLO WORLD\0} in memory ('\0' is how %s know the end of the your string is)..

Not sure if '\0' == null will yield true.. but I doubt it