如何从输入中获取数字并打印出来？

编程问答 2022-06-10

问题描述

我正在尝试制作一个将八进制数字转换为规则整数的程序。我的代码看起来像这样。

#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;
}

结果应为：

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

但是它看起来像这样：

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

我相信问题出在第32-33行：

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

但是我不知道如何具体解决问题。

解决方法

printf：* (num >> (3 * i) & 7)的最后一个参数必须为* (num >> (3 * i) & 7)（*，而不是&）

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
        }

这会产生预期的结果。

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