How do I get the digit from my input and print it out?

Question

I'm trying to make a program that converts octal numbers into regular integers. My code looks like this.

#include <stdio.h>

int main(void) {
    unsigned int c, num = 0, ct = 0;

    printf("Please input a positive octal integer and end with pressing Enter:\n");

    // Read the octal string, at most 10 characters.
    while ((c = getchar()) != '\n' && ((c >= '0' && c <= '9') && ct++ < 11)) {
        // Convert the input string to an value storing in int
        num = num << 3 | (c - '0');
    }
    // If the input is not valid, output the error message.
    if (c != '\n') {
        printf("ERROR: the input should be an octal string containing 0-7, with length less than 11!\n");
    } else {    // Output the conversion table.
        printf("i\t8^i\tdigit\tproduct\n");
        for (int i = 0; i < ct; i++) {
            printf("%u\t%u\t%u\t%u\n",
                   i, // Position i
                   1 << (3 * i), // Get 8 ** i
                   num >> (3 * i) & 7,    // Get bit at position i
                   (1 << (3 * i)) * (num >> (3 & i) & 7));    // Multiply 8 ** i to the bit at position i
        }

        // Output the decimal value
        printf("Decimal value: %d\n", num);
    }

    return 0;
}

The result should be this:

Please input a positive octal integer and end with pressing Enter:
7326
i       8^i     digit   product
0       1       6       6
1       8       2       16
2       64      3       192
3       512     7       3584
Decimal value: 3798

But instead it looks like this:

Please input a positive octal integer and end with pressing Enter:
7326
i       8^i     digit   product
0       1       6       6
1       8       2       24
2       64      3       320
3       512     7       1024
Decimal value: 3798

I believe the problem lies in line 32-33:

num >> (3 * i) & 7,    // Get bit at position i
(1 << (3 * i)) * (num >> (3 & i) & 7));    // Multiply 8 ** i to the bit at position i

But I don't know how to solve the problem specifically.

Read [this article](https://ericlippert.com/2014/03/05/how-to-debug-small-programs/) for tips on debugging your code. To check if the problem is on line 21, break the expression into smaller pieces and see if each piece gives the value you expect. — Code-Apprentice, Oct 07 '20 at 00:55
`c <= '11'` this doesn't do what you think. `'11'` is a [multicharacter literal](https://stackoverflow.com/q/3960954/995714) and not the value 11 — phuclv, Oct 07 '20 at 01:46
`num >> (3 & i) & 7` --> sure about that? It does not look like the line before `num >> (3 * i) & 7` — chux - Reinstate Monica, Oct 07 '20 at 02:05
While it is *very likely to work here*, it should be `int c`, not `unsigned c`. — Antti Haapala -- Слава Україні, Oct 07 '20 at 05:14

score 0 · Answer 1 · answered Oct 07 '20 at 02:21

The last arg of the printf : * (num >> (3 * i) & 7) needs to be * (num >> (3 * i) & 7) (*, not &)

printf("%u\t%u\t%u\t%u\n",
                   i, // Position i
                   1 << (3 * i), // Get 8 ** i
                   num >> (3 * i) & 7,    // Get bit at position i
                   (1 << (3 * i)) * (num >> (3 * i) & 7));    // Multiply 8 ** i to the bit at position i
        }

This yields your expected result.

How do I get the digit from my input and print it out?

1 Answers1