Query regarding getting input from user in C

Question

Here I am referring to a question from "codechef.com". Here this below code is used to read the initial count from the user. This will return an integer value.

This is similar to doing a scanf("%d", &n);. But most people are using this type of method to get the information from the user.

I do not understand one line in this code, and I do not understand where the character input gets converted to an integer.

int readuint()
{
int n = 0;
char c = fgetc(stdin);
do {
n = n * 10 + (c - '0');
} while ((c = fgetc(stdin)) != '\n');
return n;
}

The item in question is this line - n = n * 10 + (c - '0'); What is this line doing ??

For full code visit : http://www.codechef.com/viewsolution/1221364

Community · Accepted Answer · 2017-05-23 12:05:26Z

4

If the input is 123, then on each iteration of the loop, it calculates:

n =  0 * 10 + ('1' - '0');
n =  1 * 10 + ('2' - '0');
n = 12 * 10 + ('3' - '0');
assert(n == 123);

The character codes for the digits are always consecutive, so '1' - '0' is 1, etc.

It converts the digits of a number into a number. Get used to the idiom; you will see it a lot in C code.

The code shown is sloppy in a variety of ways:

int readuint()
{
    int n = 0;
    char c = fgetc(stdin);
    do {
        n = n * 10 + (c - '0');
    } while ((c = fgetc(stdin)) != '\n');
    return n;
}

The name indicates it is reading an unsigned integer (uint), but the type used is a signed int. The type of c should be int because fgetc() (and getc() and getchar()) return an int and not a char. There's no way to indicate that it encountered EOF. There's no protection against overflow. There's no protection against non-digits in the input. Fixing all those requires quite a lot of code, but basic self-protection for the code means it should be more like:

int readint(void)
{
    int n = 0;
    int c;
    while ((c = fgetc(stdin) != EOF && isdigit(c))
        n = n * 10 + (c - '0');
    if (c != EOF && c != '\n')
        ungetc(fp, c);
    return n;
}

There's still no protection against overflow, but it has rudimentary protection against EOF and non-digits in the input (leaving characters other than a newline or EOF to be reprocessed by putting it back for the next read operation).

'1'-'0' (char) = (int)1? How does this conversion happen? From char to int: is it because we are assigning to an integer container — i.e int n?

As Elchonon Edelson said, the character constants such as '0' and '1' are integer constants in C (they'd be char constants in C++), and the values for '0' and '1' are very often 48 and 49 respectively (but the C standard does not guarantee that!). So, 49 - 48 gives 1, of course.

edited May 23, 2017 at 12:05

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answered Jun 27, 2013 at 0:13

Jonathan Leffler

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6 Comments

Bala Over a year ago

'1'-'0' (char) = (int)1 ??? how this conversion is happening ?? from char to int. is it because we are assigning to an integer container ? i.e int n ??

This isn't my real name Over a year ago

The character constants (such as '0') actually have int type in C. Additionally, if you assign a value from a smaller-sized integer to a variable of a larger integer type, the value is promoted automatically. It doesn't require effort to convert a 1 that is of char type or short type to a 1 of int type or long type...

chux Over a year ago

Would not scanf() functionality also ungetc('\n')? BTW: I believe it is OK to ungetc(EOF) C11 7.21.7.10.4 "If the value of c equals that of the macro EOF, the operation fails and the input stream is unchanged." Thus recommend dropping the if (c != EOF && c != '\n').

chux Over a year ago

Further simplification idea: change to while (isdigit(c = fgetc(stdin))). isdigit(EOF) returns 0.

Jonathan Leffler Over a year ago

@chux: I thought about that when writing the code and decided I didn't like it and wouldn't write it like that. You're correct; it could be done. However, I don't feel that it is good style, for all it is correct.

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