Pointers & Arrays Pointing Array issue

Question

Firstly, Sorry about my bad english. I wanna ask something that I expect amazing. I'm not sure this is amazing for everyone, but It is for me :) Let me give example code

char Text[9] = "Sandrine";
for(char *Ptr = Text; *Ptr != '\0'; ++Ptr)
cout << Ptr << endl;

This code prints

Sandrine
andrine
ndrine
drine
rine
ine
ne
e

I know it's a complicated issue in C++. Why İf I call Ptr to print out screen it prints all of array. However if Text array is a dynamic array, Ptr prints only first case of dynamic array(Text). Why do it happen? Please explain C++ array that how it goes for combination of pointing array.

thanks for helping.

Because there's a special output handler for const char *. — Qaz
– Qaz, Commented Nov 9, 2013 at 22:58
@chris did you mean that Text is const char that point all of arrays elements? — burakim
– burakim, Commented Nov 9, 2013 at 23:01
new char("play") doesn't create an array! Instead, it should fail to compile: there is no way to construct a char from a char const*. — Dietmar Kühl
– Dietmar Kühl, Commented Nov 9, 2013 at 23:06

Dietmar Kühl · Accepted Answer · 2013-11-09 23:02:20Z

3

There is nothing particular special about arrays here. Instead, the special behavior is for char const*: in C, pointers to a sequence of characters with a terminating null characters are used to represent strings. C++ inherited this notion of strings in the form of string literals. To support output of these strings, the output operator for char const* interprets a pointer to a char to be actually a pointer to the start of a string and prints the sequence up to the first null character.

answered Nov 9, 2013 at 23:02

Dietmar Kühl

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AndersK · Accepted Answer · 2013-11-09 23:09:02Z

0

When you write

char Text[9] = "Sandrine";

the "Text" is an address in memory, it is the starting address of your string and in its first location there is a 'S' followed by the rest of the characters. A string in C is delimited by a \0 i.e. "S a n d r i n e \0"

When you write

for(char *Ptr = Text; *Ptr != '\0'; ++Ptr)
  cout << Ptr << endl;

when the for loop runs the first time it prints the whole string because Ptr points to the start of the string char* Ptr = Text when you increment Ptr you are pointing to the next character Text + 1 i.e. 'a' and so on once Ptr finds \0 the for loop quits.

answered Nov 9, 2013 at 23:09

AndersK

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