不返回字符串。该程序将 123 之类的数字转换为 "One Two Three" 之类的单词，为什么最后我没有得到任何东西？

Question

#include<stdio.h>
#include<string.h>

char *int_to_string( int n );

void main()
{
    int n;

    printf("Enter the number : ");
    scanf("%d",&n);

    printf("\n%d in words is : %s.\n",n,int_to_string(n));
}

char *int_to_string( int n)
{
    char str[100]="";

    if(n<10)
    {
          switch(n)
        {
            case 0: return "Zero";
            case 1: return "One";
            case 2: return "Two";
            case 3: return "Three";
            case 4: return "Four";
            case 5: return "Five";
            case 6: return "Six";
            case 7: return "Seven";
            case 8: return "Eight";
            case 9: return "Nine";
        }
    }

    else
    {
        strcat(str,int_to_string(n/10));
        strcat(str," ");

        return strcat(str,int_to_string(n%10));
    }
}

函数int_to_string() 应该return 一个字符串，其中包含传递的单词中的数字等价物。它适用于单个数字（即 0-9 ）但高于它就什么也没有。

Answer 1

函数有未定义的行为。

它returns一个指向局部数组str的指针，通常在退出函数后被销毁。

考虑到最好将参数定义为 unsigned int 类型。否则函数需要检查数字是否不为负数。

您可以通过声明第二个参数来简化任务，该参数将指定一个以零结尾的字符数组来存储结果字符串。

或者你必须动态分配内存。

这里展示了这两种方法。

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

char * int_to_string( unsigned int n )
{
    if( n < 10 )
    {
        char *p = "";

        switch( n )
        {
        case 0: 
            p = "Zero";
            break;
        case 1: 
            p = "One";
            break;
        case 2: 
            p = "Two";
            break;
        case 3: 
            p = "Three";
            break;
        case 4: 
            p = "Four";
            break;
        case 5: 
            p = "Five";
            break;
        case 6: 
            p = "Six";
            break;
        case 7: 
            p = "Seven";
            break;
        case 8: 
            p = "Eight";
            break;
        case 9: 
            p = "Nine";
            break;
        }
        char *q = malloc( strlen( p ) + 1 );
        strcpy( q, p );
        free( p );
        return q; 
    }
    else
    {
        char *q = int_to_string( n / 10 );
        char *p = int_to_string( n % 10 );

        q = realloc( q, strlen( q ) + strlen( p ) + 2 );
        strcat( q, " " );
        return strcat( q, p );
    }
}

char * int_to_string1( unsigned int n, char *s )
{
    if( n < 10 )
    {
        char *p = "";

        switch( n )
        {
        case 0: 
            p = "Zero";
            break;
        case 1: 
            p = "One";
            break;
        case 2: 
            p = "Two";
            break;
        case 3: 
            p = "Three";
            break;
        case 4: 
            p = "Four";
            break;
        case 5: 
            p = "Five";
            break;
        case 6: 
            p = "Six";
            break;
        case 7: 
            p = "Seven";
            break;
        case 8: 
            p = "Eight";
            break;
        case 9: 
            p = "Nine";
            break;
        }

        return strcat( s, p );
    }
    else
    {
        strcat( int_to_string1( n / 10, s ), " " );

        return int_to_string1( n % 10, s );
    }
}

int main( void )
{
    unsigned int n = 1234567890;
    char *s = int_to_string( n );

    puts( s );

    free( s );

    char s1[100];
    s1[0] = '[=10=]';

    puts( int_to_string1( n, s1 ) );
}

程序输出为

One Two Three Four Five Six Seven Eight Nine Zero
One Two Three Four Five Six Seven Eight Nine Zero

Answer 2

请在发送最终字符串之前使用字符串复制。在你的代码中添加最后两行它会起作用。

    char *int_to_string( int n)
    {
       char str[100]="";
       char str1[100]="";//
       .
       .
       .
       strcat(str,int_to_string(n/10));
       strcat(str," ");     
       strcat(str,int_to_string(n%10));
       strcpy(str1,str);//create one more str1 array of 100 and copy final data
       return str1;// return str1 array of data
     }

Answer 3

1)

在你的函数str作为堆栈上的自动变量在int_to_string返回后被销毁。但是你需要 str 才能存活更多 int_to_string 个电话！所以你必须在调用之间保留 str。

2)

case 0: return "Zero";
....

上面的代码在递归调用中将无法正常工作，单词"Zero"必须添加到str字符串

     case 0: strcat(str,"Zero"); return str;

但是为什么要忍受递归调用呢？递归可以用简单的循环代替。显示了两种解决方案。

#include <stdio.h>
#include <stdlib.h> 
#include <ctype.h>
#include <string.h>
#include <math.h>

void print_digit(int digit)
{
    switch(digit)
    {
        case '0': printf("Zero "); break;
        case '1': printf("One ");  break;
        case '2': printf("Two ");  break;
        case '3': printf("Three ");break;
        case '4': printf("Four "); break;
        case '5': printf("Five "); break;
        case '6': printf("Six ");  break;
        case '7': printf("Seven ");break;
        case '8': printf("Eight ");break;
        case '9': printf("Nine "); break;
   }
}

char * return_digit_word(int digit)
{
    switch(digit)
    {
        case '0': return("Zero "); break;
        case '1': return("One ");  break;
        case '2': return("Two ");  break;
        case '3': return("Three ");break;
        case '4': return("Four "); break;
        case '5': return("Five "); break;
        case '6': return("Six ");  break;
        case '7': return("Seven ");break;
        case '8': return("Eight ");break;
        case '9': return("Nine "); break;
   }
}

char *int_to_string(int n,char str[],char numStr[]) 
{
     if(n<10)
     {
        switch(n)
        {
            case 0: strcat(str,"Zero");break;
            case 1: strcat(str,"One");break;
            case 2: strcat(str,"Two");break;
            case 3: strcat(str,"Three");break;
            case 4: strcat(str,"Four");break;
            case 5: strcat(str,"Five");break;
            case 6: strcat(str,"Six");break;
            case 7: strcat(str,"Seven");break;
            case 8: strcat(str,"Eight");break;
            case 9: strcat(str,"Nine");break;
        }
        return str;
    }
    else{
        int digit = numStr[0]-'0';
        int newNr =  n - digit*pow(10,strlen(numStr)-1);        

        strcat(str, return_digit_word(numStr[0]));
        sprintf(numStr, "%d", newNr); 

        return int_to_string(newNr,str,numStr);
    }
} 


int main(void) {
    int n,i;
    char str[100]="";
    char numStr[100]="";

    n = 1234567890;

    sprintf(numStr, "%d", n);
    printf("\n%d in words is : %s\n",n, int_to_string(n,str,numStr) );

    printf("\n%d in words is : ",n);
    sprintf(numStr, "%d", n);
    for(i=0;i<strlen(numStr);i++)
    {
       print_digit(numStr[i]);
    }

   return 0;
}

n=1234567890 的输出：

1234567890 in words is : One Two Three Four Five Six Seven Eight Nine Zero

1234567890 in words is : One Two Three Four Five Six Seven Eight Nine Zero

不返回字符串。该程序将 123 之类的数字转换为 "One Two Three" 之类的单词，为什么最后我没有得到任何东西？

Not returning string. This program converts digit like 123 to words like "One Two Three", why in the end I am not getting anything?

c

function

recursion

strcat