Plzz explain (*num !='\o')

Question

from line 41 to 78
https://code.sololearn.com/cJza6635d7Ir/?ref=app

Answer 1

It is '\0', the null character.

C-style strings, i.e. const char* are null-terminated. The value held by a C-style string is a pointer pointing to the first character in the character array. The last character in the array is '\0'.

Doing:

while (*num != '\0')

would mean that the loop is executed until the end of the string is reached.

Answer 2

Well, that's another story. We have to look into the ASCII value of the character '0'.

The character '0' has an ASCII value of 48.

http://www.asciitable.com/

*num returns a character value pointed by the address stored within num. When we want to convert a numeric character value to it's integer counterpart, it is a norm to minus that character ASCII by 48 (i.e. '0').

Hence, character '1' of ASCII value 49, will be converted to

'1' - '0'
49 - 48
1

What "if (*num -'0' != 0)" literally means, is:

"If the numeric character pointed by num is not 0".

However (!), I believe the notation is redundant here, since we could have easily achieved the same thing via:

if (*num != '0')