使用C编程语言K＆R的atoi函数

Question

I quite don't know what this loop does.

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;
}

This part I don't get:

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

I get the whole for loop inside parentheses, and what s[i] - '0' does. But I don't get what kind of operation is going on here --> n = 10 * n .

I don't know what n is representing and why is multiplying 10. I know it's converting string of digits to numeric equivalent, but I just don't get the whole operation there.

Answer 1

But I don't get what kind of operation is going on here --> n = 10 * n

That's just how you build a number digit by digit. It's basically the same as it would work if you were writing a calculator. If I wrote a simple calculator, here's how it would handle the input 5 4 7 :

1. Start with 0
2. 5 ==> 0*10 + 5 = 5
3. 4 ==> 5*10 + 4 = 54
4. 7 ==> 54*10 + 7 = 547

Basically, atoi does, the exact same thing, but instead of reading each digit from button presses, it's reading them from a string. Each time you read a new digit, you do n *= 10 to make room for that next digit, which just gets directly added on the end.

Answer 2

n is the digits that has already been processed. For instance, for the string "123" , first, the program gets digit 1 , convert it to integer and store it in n , and then get the next digit 2 , this is where n = 10 * n is useful, the previous 1 is multiplied by 10 , and added to the next digit 2 , the result is 12 , and this is stored as the current n .

The same goes on, when processing 3 , the previous stored 12 is multiplied by 10 , results in 120 and added to 3 , ended as 123 as result.