Question

In the given program:

int main() {
  char *p = "\0777";
  printf("%d %d %d\n",p[0],p[1],p[2]);
  printf("--%c-- --%c-- --%c--\n",p[0],p[1],p[2]);
  return 0;  
}

It is showing the output as:

63 55 0
–?– –7– —-

I can understand that it is converting the first two characters after \0 (\077) from octal to decimal but can any one explain me why 2 characters, why not 1 or 3 or any other ?

Please explain the logic behind this.

Answer 1

char *p = "\07777";

Here a string literal assigned to a pointer to a char.

"\07777"

In this string literal octal escape sequence is used so first three digits represents a octal number.because rules for octal escape sequence is—

You can use only the digits 0 through 7 in an octal escape sequence. Octal escape sequences can never be longer than three digits and are terminated by the first character that is not an octal digit. Although you do not need to use all three digits, you must use at least one. For example, the octal representation is \10 for the ASCII backspace character and \101 for the letter A, as given in an ASCII chart.

SO your string literal stored in memory like

1st byte as a octal number 077 which is nothing but 63 in decimal and ‘?’ in character

2nd and 3rd byte as a characters ‘7’ and ‘7’ respectively

and a terminating character ‘\0’ in last.

so your answer are as expected 1st,2nd,3d byte of the string literal.

for more explanation you can visit this web site

http://msdn.microsoft.com/en-us/library/edsza5ck.aspx