用C中的字符串替换整数

Question

我编写的代码用strings 替换了从0 到3 的整数。我只被允许使用getchar() 和putchar()。如果input 为1，则输出将变为"one"。

#include <stdio.h>

int main()
{
    int c;            
    char* arr[4] = {"zero", "one", "two","three"};
    int i;

        while ((c = getchar ()) != EOF) 
        {
            if(c==0+'0') {
                char* str = arr[0];
                for (i = 0; str[i] != '\0'; i++) {
                    putchar(str[i]);
                }
            }
            else if(c==1+'0') {
                char* str = arr[1];
                for (i= 0; str[i] != '\0';i++) {
                    putchar(str[i]);
                }
            }
            else if(c==2+'0') {
                char* str = arr[2];
                for (i = 0; str[i] != '\0'; i++) {
                    putchar(str[i]);
                }
            }
            else if(c==3+'0') {
                char* str = arr[3];
                for (i = 0; str[i] != '\0'; i++) {
                    putchar(str[i]);
                }   
            }
            else
                putchar(c);
   }

return 0;
}

code 很长。有没有更短的写法？

如果我输入33，输出将是"threethree"。谁能给我建议我如何修改我的代码不这样做？

注意：我也不能使用函数。

Answer 1

你可以用一个变量来记住最后的输入和比较，这样你就不会打印连续的字符了。

#include <stdio.h>

int main()
{
    int c;
    char* arr[4] = {"zero", "one", "two","three"};
    int i;
    char last_input = '9';

    while ((c = getchar ()) != EOF)
    {
        if(c != last_input && '0' <= c  && c <= '3') {
            last_input = c;
            int index = c - '0';
            char* str = arr[index];
            for (i = 0; str[i] != '\0'; i++) {
                putchar(str[i]);
            }
        }
        else{
            putchar(c);
        }

    }
    return 0;
}

Answer 2

您可以使用这样的一个 if 条件来压缩您的 if 语句：

#include <stdio.h>

int main()
{
    int c;            
    char* arr[4] = {"zero", "one", "two","three"};
    int i;

    while ((c = getchar ()) != EOF) {
    int k = c-'0';
        if(k>=0 && k<=3) {
            char* str = arr[k];
            for (i= 0; str[i] != '\0';i++) {
                putchar(str[i]);
            }
        }
        else {
            putchar(c);
        }
    }
return 0;
}

Answer 3

这是完成相同任务的简单方法。我试图解释 cmets 中的逻辑。

int main(void) {
        char *arr[11] = {"zero", "one", "two","three","four","five","six","seven","eight","Nine","Ten"};
        int *input = malloc(sizeof(*input))/*1st time 4 byte */ , row = 1;
        while( (input[row-1] = getchar())!=EOF ) {

                if(input[row-1]==10) /* if ENTER key is presed */
                        break;
                input[row-1] = input[row-1] - 48;/* convert it */
                printf("%s ",arr[ input[row-1]%10 ]);/* its simple, just think on it */
                row++;
                input = realloc(input,row * sizeof(*input));/* reallocate based on number of input */
        }
        /* free dynamically allocated memory @TODO*/
        return 0;
}

我只是给出了提示，让它通用，比如输入小于零时写条件等。我希望它有所帮助。

Answer 4

这里我的代码使用循环来缩短你的代码。

#include <stdio.h>

int main()
{
    int c;            
    char* arr[4] = {"zero", "one", "two","three"};
    int i, j;

    while ((c = getchar ()) != EOF) 
    {
        for(j = 0; j < 4; j++)
        {
            if(c == j + '0')
            {
                char* str = arr[j];
                for (i = 0; str[i] != '\0'; i++)
                {
                    putchar(str[i]);
                }
                j = 10; // just to detect processed character
                break;
            }
        }
        if(j != 10)
        {
            putchar(c);
        }
    }
    return 0;
}