pointer arithmetic on arrays

Question

When I run the code below my output is not what I expect.

My way of understanding it is that ptr points to the address of the first element of the Str array. I think ptr + 5 should lead to the + 5th element which is f. So the output should only display f and not both fg.

Why is it showing fg? Does it have to do with how cout displays an array?

#include <iostream>
using namespace std;

int main()
{
    char *ptr;
    char Str[] = "abcdefg";

    ptr = Str;
    ptr += 5;

    cout << ptr;

    return 0;
}

Expected output: f

Actual output: fg

ptr - pointer to string. If you want to print char, use *ptr: cout << *ptr; — SashaMN
– SashaMN, Commented Jan 17, 2016 at 21:46
Replace "lead" with more appropriate "point to" and you'll get the idea. — Drop
– Drop, Commented Jan 17, 2016 at 21:52

user007 · Accepted Answer · 2016-01-17 22:00:24Z

6

When you declare:

char Str[] = "abcdefg"

The string abcdefg is stored implicitly with an extra character \0 which marks the end of the string.

So, when you cout a char* the output will be all the characters stored where the char * points and all the characters stored in consecutive memory locations after the char* until a \0 character is encountered at one of the memory locations! Since, \0 character is after g in your example hence 2 characters are printed.

In case you only want to print the current character, you shall do this ::

cout << *ptr;

edited Jan 17, 2016 at 22:00

answered Jan 17, 2016 at 21:42

user007

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Ziezi · Accepted Answer · 2016-01-17 22:06:07Z

2

Why is it showing fg?

The reason why std::cout << char* prints the string till the end instead of a single char of the string is , because std::cout treats a char * as a pointer to the first character of a C-style string and prints it as such.¹

Your array:

char Str[] = "abcdefg";

gets implicitly assigned an '\0'at the end and it is treated as a C-style string.

Does it have to do with how std::cout displays an array?

This has to do with how std::cout handles C-style strings, to test this change the array type to int and see the difference, i.e. it will print a single element.

^{1. This is because in C there are no string types and strings are manipulated through pointers of type char, indicating the beginning and termination character: '\0', indicating the end.}

edited Jan 17, 2016 at 22:06

answered Jan 17, 2016 at 21:51

Ziezi

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