I don't understand atoi function in K&R C book?

Question

#include <stdio.h>
#include <string.h>

int main(void)
{
    char s[]= "9";
    printf("atoi = %d",atoi(s));
    system("pause");
    return 0;
}

int atoi(char s[])
{
    int i=0,n=0;

    for(i;s[i]>='0' && s[i]<='9';i++)
        n=10*n + (s[i]-'0');
    return n;

}

In above program it gave me result 9 as per program it should print ascii value for 9 and I don't understand what this for loop does.

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

Answer 1

Lets break this down:

for (i;

This creates a for loop, with the loop variable i. This is not necessary, but more of a coding style.

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

This checks to make sure that the character at that index is inside the range for a decimal character (0 - 9), and if it is not, it exits the loop, then returns the number.

i++

After the loop runs, this increases the index you are checking in the string by one.

n = 10 * n

This adds an extra digit to 'n' by multiplying by 10, because you know that if you have one more character in your number, it must be multiplied by ten (say I start parsing 100, I read the first two strings, and have 10, there is one more character, so I multiply by ten to get 100.

+ (s[i]-'0');

This adds the next digit to 'n', the result, which is determined by subtracting the character at the current index by '0', which, when in the range of 0 - 9, returns the integer for that number (if this confuses you, take a look at an ASCII Chart.

Hopefully this helped you understand.