逐个字符读取-总是到达无效字符

Question

我是C语言和编程世界的新手。我被要求以20为基数得到一个反转的数字，并打印出他转换成十进制的结果。我知道这不是最有效的代码，但这是我目前所学到的最好的代码。当我运行这个程序并输入一个数字时，我总是得到默认的情况...我遗漏了什么？谢谢您:)

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
int main() {
    char x;
    int exponent= 1;
    int sum = 0;
    int flag = 0;
    printf("Enter a reversed number is base 20\n");
    scanf(" %c", &x);
    while (x != "\n") {
        switch (x) {
        case '0':sum += 0;
            break;
        case '1':sum += 1 * exponent;
            break;
        case '2':   sum += 2 * exponent;
            break;
        case '3':sum += 3 * exponent;
            break;
        case '4':sum += 4 * exponent;
            break;
        case '5':sum += 5 * exponent;
            break;
        case '6':sum += 6 * exponent;
            break;
        case '7':sum += 7 * exponent;
            break;
        case '8':sum += 8 * exponent;
            break;
        case '9':sum += 9 * exponent;
            break;
        case 'A':
        case 'a': sum += 10 * exponent;
            break;
        case 'B':
        case 'b':sum += 11 * exponent;
            break;
        case 'C':
        case 'c' : sum += 12 * exponent;
            break;
        case 'D':
        case 'd': sum += 13 * exponent;
            break;
        case 'E':
        case 'e': sum += 14 * exponent;
            break;
        case 'F':
        case 'f': sum += 15 * exponent;
            break;
        case 'G':
        case 'g': sum += 16 * exponent;
            break;
        case 'H':
        case 'h': sum += 17 * exponent;
            break;
        case 'I':
        case 'i': sum += 18 * exponent;
            break;
        case 'J':
        case 'j': sum += 19 * exponent;
            break;
        default:flag++;
            break;
        }
        if (flag == 1) {
            printf("Error! %c is not a valid digit in base 20", x);
            break;
        }
        else {
            exponent *= 20;;
            scanf("%c", &x);
        }
    }
    if (flag == 1)
        return 0;
    else
        printf(sum);
    return 0;
}

Answer 1

你真的应该看看你的compiler's warnings...：

<source>: In function 'main':
<source>:10:14: warning: comparison between pointer and integer
   10 |     while (x != "\n") {
      |              ^~
<source>:10:14: warning: comparison with string literal results in unspecified behavior [-Waddress]
<source>:77:16: warning: passing argument 1 of 'printf' makes pointer from integer without a cast [-Wint-conversion]
   77 |         printf(sum);
      |                ^~~
      |                |
      |                int
In file included from <source>:2:
/usr/include/stdio.h:332:43: note: expected 'const char * restrict' but argument is of type 'int'
  332 | extern int printf (const char *__restrict __format, ...);
      |                    ~~~~~~~~~~~~~~~~~~~~~~~^~~~~~~~

您正在将x与一个地址进行比较-字符串常量"\n“的地址(位于程序内存空间的某一部分)。您可能是想将x与'\n‘进行比较。

此外，printf()函数在参数之前接受一个格式字符串，因此它应该是printf("%d\n", sum); (为了打印换行符)。

PS 1:总是在编译代码时打开更多的警告，例如gcc -W -Wall (也许还有其他警告标志)，以便更好地捕获这些类型的打字错误和小错误。

PS 2:正如@Jabberwocky指出的那样，你的程序可以变得更短(不会失去可读性)；另一方面，你应该总是检查scanf()和类似函数的返回值，这些函数可能会因为用户的输入而失败：

#include <stdio.h>
#include <stdlib.h>

int main() {
    char x;
    int exponent = 1;
    int sum = 0;
    int num_scanned;
    printf("Enter a reversed number is base 20\n");
    num_scanned = scanf(" %c", &x);
    if (num_scanned != 1) {
        fprintf(stderr, "Invalid input\n");
        exit(EXIT_FAILURE);
    }
    while (x != '\n') {
        int digit;
        if      (x >= '0' && x <= '9') { digit = x-'0'; }
        else if (x >= 'a' && x <= 'j') { digit = x-'a'; }
        else if (x >= 'A' && x <= 'J') { digit = x-'A'; }
        else {
            fprintf(stderr, "%c is not a valid digit in base 20.", x);
            exit(EXIT_FAILURE);
        }
        sum += digit * exponent;
        exponent *= 20;
        num_scanned = scanf("%c", &x);
        if (num_scanned != 1) {
           fprintf(stderr, "Invalid input\n");
           exit(EXIT_FAILURE);
        }
    }
    printf("%d\n",sum);
    return EXIT_SUCCESS;
}