在C语言中将“字符”转换为“整数”的逻辑

Question

everyone!

please help me understand the following problem... So i will have a STRING-type input of a note, looks like "A5" or "G#2" or "Cb4" etc. And i need to extract an octave index, which is the last digit "5" or "2" or "4"... And after exctraction i need it as an int-type.

So I did this:

string note = get_string("Input: ");
int octave = atoi(note[strlen(note) - 1]);
printf("your octave is %i \n", octave);

But it gave me an error "error: incompatible integer to pointer conversion passing 'char' to parameter of type 'const char *'; take the address with & [-Werror,-Wint-conversion]"

Then I tryied to throug away the math from the function, and did this:

int extnum = strlen(note) - 1;
int octave = atoi(note[extnum]);

It didn't work as well. So i did my reserch on atoi function and i don't get it...

1. ATOI expects a string (CHECK)
2. Converts it to an interger, not the ASCII meanning (CHECK)
3. Library for atoi function (CHECK)

What I am doing in basically asking "take n-th character of that string and make it an int".

After googling for some time a found an other code example where a guy uses atoi with this symbol '&'. So i did this :

int octave = atoi(&note[strlen(note) - 1]);

And IT WORKED! But I can't understand WHY it worked with the & symbol and didnt work without it....Cause it always worked without it! There was a million times i was giving a single-character string like '5' or so ond just used atoi and it worked perfectly...

Plesase help me, why in this case it acts so weird?

Answer 1

C does not have a native string type. Strings are usually represented as char array or a pointer to char.

Assuming that string is just a typedef to char * .

if note is an array of chars, note[strlen(note)-1] is just the last character. Since atoi expects a pointer to char (which has to be null-terminated) you have to pass the address of the char and not the value.

The task to convert one char digit to int could also be solved easier:

int octave = note[strlen(note) - 1] - '0';

Answer 2

The function atoi takes a pointer to a character array as the input parameter ( const char* ). When you call note[strlen(note) - 1] this is a single character ( char ), in order to make atoi work you need to provide the pointer. You do that by adding & as you've done. This then works, because right after that single digit there is a null character \\0 that terminates the string - because your original string was null-terminated.

Note however that doing something like this would not be a good idea:

char a = '7';
int b = atoi(&a);

as there is no way to be sure what the next byte in memory is (following the byte that belongs to a ), but the function will try to read it anyway, which can lead to undefined behaviour.

Answer 3

The last character is... well a character! not a string. So by adding the & sign, you made it a pointer to character ( char* )!

You can also try this code:

char a = '5';
int b = a - '0';

Gives you ASCII code of 5 minus ASCII code of 0

Answer 4

From the manual pages , the signature of atoi is: int atoi(const char *nptr); . So, you need to pass the address of a char .

When you do this: atoi(note[strlen(note) - 1]) you pass the char itself. Thus, invoking UB .

When you use the & , you are passing what the function expects - the address. Hence, that works.

Answer 5

atoi excepts a string (a pointer to character), not a single character.

However, you should never use atoi since that function has bad error handling. The function strtol is 100% equivalent but safer.

You need to do this in two steps:

1. Find the first digit in the string.
2. From there, convert it to integer by calling strtol .

1) can is solved by looping through the string, checking if every item is a digit by calling isdigit from ctype.h. Simultaneously, check for the end of the string, the null terminator \\0 . When you find the first digit, save a pointer to that address.

2) is solved by passing the saved pointer to strtol , such as result = strtol(pointer, NULL, 10); .