在C中有Atoi函数吗？

Question

int atoi(char* s)
{
   int i,n;
   n=0;
   for (i=0; (s[i]>='0') && (s[i]<='9'); i++)
       n = 10 * n + (s[i] - '0');

   return n;
}

In this code what is s[i]-'0' doing? Can anyone please explain the detailed working of this function?

Answer 1

Have a look at the table in the link below-

http://www.asciitable.com/

The table is called ASCII Table and it is one of the character-encoding schemes used to represent characters in the binary world.

You can see the decimal numbers 0-9 are represented by numbers 48-57 in the ASCII table. All digits(0-9) are stored as characters.

If your computer stores 48 for decimal number 0, 49 for decimal number 1, 50 for decimal number 2 and so-on. Then, to convert a ASCII number into decimal number, you just need to subtract 48 from ASCII number. For example,

48 - 48 => 0

49 - 48 => 1

50 - 48 => 2

.. and so-on

'0' also represents 48. It is a character form of number 48. That's why, the equation n = 10 * n + (s[i] - '0'); has '0'.

Answer 2

In this code what is s[i]-'0' doing?

In C, each character like '0' , 'A' , '+' , ' ' is assigned a numeric value or code. C requires that the codes for '0' , '1' , '2' ... '9' are sequential but does not specify their values.

When code performs the below test, it knows that s[i] has a value within codes '0' and '9' . Since these codes are sequential, the only values s[i] could have are '0' , '1' , '2' ... '9' .

(s[i]>='0') && (s[i]<='9')

By subtracting '0' from s[i] , code obtains the difference:

`0`-'0' --> 0
`1`-'0' --> 1
`2`-'0' --> 2
...
`9`-'0' --> 9

Code has successfully translated the character code for an numeric character into a corresponding integer value.

Answer 3

In your function, s is a sequence of ASCII encoded characters: a string. _{Actually, the encoding does not matter as long as it has the characters 0 through 9 as a sequence.} Let's say it has the value "123ABC" , for this explanation.

n is the output number. At the start of the function it is initialized to zero.

In the first iteration of the loop we have

i=0
s[i] = '1'  (which is encoded as 49, in ASCII)
n = 0

so the math is like this:

n = n * 10 + (s[i] - '0')
n = 0 * 10 + ('1' - '0')
n = '1' - '0'

And converting the character syntax to the ASCII encoding gives this:

n = 49 - 48
n = 1

In the next iteration we have:

i = 1
s[i] = '2' (ASCII 50)
n = 1

n = n * 10 + (s[i] - '0')
n = 1 * 10 + ('2' - '0')
n = 10 + ('2' - '0')
n = 10 + (50 - 48)
n = 10 + 2
n = 12

And, in the third iteration:

i = 2
s[i] = '3' (ASCII 51)
n = 12

n = n * 10 + (s[i] - '0')
n = 12 * 10 + ('3' - '0')
n = 120 + ('3' - '0')
n = 120 + (51 - 48)
n = 120 + 3
n = 123

And, in the final iteration s[i] = 'A' which is not in the range specified by the if statement, so the loop exits.

As you can see it has correctly converted the string "123" to the number 123 .