what does string - '0' do (string is a cha

问题:

what does this do

while(*string) {
    i = (i << 3) + (i<<1) + (*string -'0');
    string++;
}

the *string -'0'

does it remove the character value or something?

回答1:

This subtracts from the character to which string is pointing the ASCII code of the character '0'. So, '0' - '0' gives you 0 and so on and '9' - '0' gives you 9.

The entire loop is basically calculating "manually" the numerical value of the decimal integer in the string string points to.

That's because i << 3 is equivalent to i * 8 and i << 1 is equivalent to i * 2 and (i << 3) + (i<<1) is equivalent to i * 8 + i * 2 or i * 10.

回答2:

It converts the ascii value of 0-9 characters to its numerical value.

ASCII value of '0' (character) is 48 and '1' is 49. So to convert 48-56('0'-'9') to 0-9, you just need to subtract 48 from the ascii value. that is what your code line [ *string -'0' ] is doing.

回答3:

Since the digits 0-9 are guaranteed to be stored contiguously in the character set, subtracting '0' gives the integer value of whichever character digit you have.

Let's say you're using ASCII:

char digit = '6'; //value of 54 in ASCII
int actual = digit - '0'; //'0' is 48 in ASCII, therefore `actual` is 6.

No matter which values the digits have in the character set, since they're contiguous, subtracting the beginning ('0') from the digit will give the digit you're looking for. Note that the same is NOT particularly true for the letters. Look at EBCDIC, for example.