接受用户的输入

Question

 i tried to take input from user 
 input type is not determined(can be char or int)
 i wanna take input and store in pointer array
 while i doing that job forr each pointer i wanna take place from leap area
 that is using malloc
  but below code doesnot work why??? 

 int main(void)
{
    char *tutar[100][20],temp;
    int i;
    int n;
    i=0;

    while(temp!='x')
    {
        scanf("%c",&temp);
        tutar[i]=malloc(sizeof(int));
        tutar[i]=temp;
        ++i;
    }

    n =i;
    for(i=0;i<=n;++i)
    {
        printf(" %c  ",*tutar[i]);
    }
    printf("\n\n");

   /*for(i=0;i<=n;++i)
   {
        printf("%d",atoi(*tutar[i]));
   }
    */
}

note that; this cite has problem when rewrite(edit) the previous mail it is general problem or not

Answer 1

There are several problems in your code, including:

• you declare tutar as a two-dimensional array of pointers to character, then use it as a one-dimensional array
• tutar[i]=temp assigns the value of temp (a char) to tutar[i] (a pointer to char), effectively overwriting the pointer to the newly reserved memory block
• you don't initialize temp , so it will have garbage value - occasionally it may have the value x , in which your loop will not execute

Here is an improved version (it is not tested, and I don't claim it to be perfect):

int main(void)
{
    char *tutar[100], temp = 0;
    int i = 0;
    int n;

    while(temp!='x')
    {
        scanf("%c",&temp);
        tutar[i]=malloc(sizeof(char));
        *tutar[i]=temp;
        ++i;
    }

    n =i;
    for(i=0;i<=n;++i)
    {
        printf(" %c  ",*tutar[i]);
    }
    printf("\n\n");
}

Note that unless you really need to allocate memory dynamically, you would be better off using a simple array of chars:

    char tutar[100], ...
    ...

    while(temp!='x')
    {
        scanf("%c",&temp);
        tutar[i++]=temp;
    }
    ...

For the sake of brevity, I incremented i within the assignment statement.