Question

作为一项学习练习，我正在创建一个程序，将通过argv[1]输入的字符转换为整数。然后程序找到整数的数字均值。

例如：abc为123，1+2+3 = 6，六为数字均值。该程序的第一部分有效，但我无法正确编码找到数字均值的部分。

word abc的输出应为123 6;相反，它是123 150。

编辑：已解决！

#include <stdio.h>
#include <strings.h>

int main(int argc, char *argv[])
{
    char stra[27], strb[0];
    int str_num, str_len;
    int final = 0;

    if (argv[1][0] < 'a')
    {
        printf("!Argument missing!");
        return 0;
    }

    strcpy(stra, argv[1]);
    str_len = strlen(stra);

    for (str_num = 0; str_num < str_len; str_num++)
    {
        if (stra[str_num] <= 'z' && stra[str_num] >= 'a')
        {
            strb[str_num] = (stra[str_num] - 'a' + 1) % 9 + '0';
            if (strb[str_num] == '0')
            {
                strb[str_num] = '9';
            }
        }
        else
        {
            printf("%s !Please use only the lower case!", stra);
            return 0;
        }
    }

    for (str_num = 0; str_num < str_len; str_num++)
    {
        final += strb[str_num] - '0';
    }

    printf("%s %i", strb, final);
    return 0;
}

Answer 1

在第一次检查时，如果这是

，您的程序似乎或多或少会起作用

strb[str_num] = (stra[str_num] - 'a') % 9 + 1

此外，strb也需要27个字符，而不是0; 0元素数组在保存非零数据量方面相当差！

您需要使用argc检查参数 - 即

if (argc != 2)
    // complain here

最后，当您打印结果时，需要将序号转回字符。将strb打印为char*不会给您123，而是^A^B^C，而不是您想要的。因此，您需要在循环中打印这些字符，通过将'0'的偏移量添加回来调整每个字符。

Answer 2

好像你已经跑过缓冲区的末尾strb：

char strb[0];

Answer 3

strb[str_num] = (stra[str_num] - 'a') % 9 - 207;

207是这里最大的问题（除了你的一个一个错误）。这一行应为：

strb[str_num] = (stra[str_num] - 'a' + 1) % 9;

Answer 4

#include <stdio.h>
#include <strings.h>
int main(int argc, char *argv[])
{
    char *stra;
    char *strb;
    int str_num;
    int str_len;
    int final;
    final = 0;
    if ( (argc >= 2 && argv[1][0] < 'a') || (argc <= 1))
    {
        printf("!Argument missing!");
        return 0;
    }
    str_len = strlen(argv[1]);

    strb = (char*) malloc((str_len+1)*sizeof(char));

    for (str_num = 0; str_num < str_len; str_num++)
    {
        if (argv[1][str_num] <= 'z' && argv[1][str_num] >= 'a')
    {
        strb[str_num] = (argv[1][str_num] - 'a'+1) % 9 + '0'; //to get string with numbers from 1 to 9
        }
        else
        {
            printf("%s !Please use only the lower case!",argv[1]);
            return 0;
        }
    }
    strb[str_len] = '\0' // put the end char of the string : mandatory to print it.
    for (str_num = 0; str_num < str_len; str_num++)
    {
        final += strb[str_num] - '0';  // count the sum of the numbers from 1 to 9 ?
    }
    printf("%s %i",strb,final);
    free(strb);
    return 0;
}

使用字符串查找数字均值

4 个答案: