How to print a char array and a certain character in it?

Question

I've been reviewing my C++ lately. But I am running into a puzzle about printing a char array. The code is below:

int n = 5;
char *array1 = new char[n];
for (unsigned int i = 0; i < n - 1; i++)
    array1[i] = (char)i;
cout << array1 << endl;
cout << array1[3] << endl;
cout << *array1 << endl;

None of the three cout lines works. Could anyone tell me why?

What is not working? Does it compile? If so, is it running? If so, what is the output? — Anon Mail, Jun 27 '17 at 13:46
The `int` value `1` (for example) is not the same as the character `'1'`. [This ASCII (the most common character encoding) table](http://en.cppreference.com/w/cpp/language/ascii) should hopefully help you understand. And with that table in front of you, think about what happens if you do e.g. `1 + '0'`. — Some programmer dude, Jun 27 '17 at 13:47
You want to convert the number represented by i to a char? if i == 0, you want to print 0? — JLev, Jun 27 '17 at 13:47
They work, you're just storing and printing non-printable characters. Sounds like you could use a [good C++ book](http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list) — NathanOliver, Jun 27 '17 at 13:47
@AnonMail Well It compiles and runs. But just gave nothing out. — Holie Chen, Jun 27 '17 at 13:59
@HolieChen try using this answer - https://stackoverflow.com/questions/4629050/convert-an-int-to-ascii-character — JLev, Jun 27 '17 at 14:02

score 1 · Accepted Answer · answered Jun 27 '17 at 13:48

array1[0] == 0. cout << array1 interprets array1 as a pointer to a NUL-terminated string, and since the very first character is in fact NUL, the string is empty.

cout << array1[3] does print a character with ASCII code 3. It's a non-printable character, not visible to a naked eye. Not sure what output you expected to see there.

score 0 · Answer 2 · answered Jun 27 '17 at 14:26

0

As a separate answer, it seems you're trying to get a string which has the following : array = "1234....(n-1)"

Try :

for (int i = 0; i<(n-1); i++)
    array1[i] = (char)i - '0';

answered Jun 27 '17 at 14:26

Shivansh Jagga

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Should be '+' I suppose? – Holie Chen Jun 27 '17 at 14:30
If its ' - ' ,this is what it does : `(ASCII value of n) - (ASCII value of 0) = n` . So if its the number 0, it becomes `(ascii of 0) - (ascii of 0) =0` itself. Please up vote incase I helped ^_^ – Shivansh Jagga Jun 27 '17 at 14:33
1

I don't think so. Because (char)i means "character with ascii code of i" but not "ascii code of character 'i' ". So it should be '+'. – Holie Chen Jun 27 '17 at 14:47

How to print a char array and a certain character in it?

2 Answers2