Get to number from an array and assign to one int

问题

I'm working on my homework and trying to get two characters which are numbers from an array for example ABC10DEF

I want to get 10 and store it in an int type.

            number_holder_1 = back[3] - '0';
            number_holder_2 = back[4] - '0';

            number = number_holder_1 * 10 + number_holder_2;

This doesn't work I don't know why.

When I print out the value of number_holder_1 it does display 1 and when I display number_holder_2 it display 0 but when I print number it just display 0

I don't know how.

UPDATE:

The number, number_holder_1 and number_holder_2 are define as int type.

the array called back that holding ABC10DEF is passing from the main.

回答1:

It appears to be an implementation mistake, because using what you have given, with some better variable names, it does work.

#include <stdio.h>
#include <ctype.h>

int letter2int(char c) {
    int n;
    if (isdigit(c)) {
        n = c - '0';
    } else {
        n = -1; /* error */
    }

    /* for debugging */
    printf("For character '%c' we get number '%d'.\n", c, n);

    return n;
}

int main(int argc, char** argv) {
    const char str[] = "ABC10DEF";
    int tens, ones;
    int result;

    tens = letter2int(str[3]);
    ones = letter2int(str[4]);

    result = tens * 10 + ones;

    printf("Result: %d\n", result);

    return 0;
}

This can be generalized to either form a atoi function (ASCII to integer) or extract the first number that occurs in a string (terminated by any non-digit character) by using a loop and a char pointer to index over the string str.

Using i as the zero-based index, result += number * (int)pow(10, i);.

来源：https://stackoverflow.com/questions/10640288/get-to-number-from-an-array-and-assign-to-one-int

标签

logic