Dynamically stored strings. What is the exact method?

[Avon]

Hello there.

I'm (kind of) new to programming as you can see, and I have a question.

I want the user to input a string (for example: a product code) but store it dynamically. There must be an other way instead of a static array with, let's say, 50 elements. What I have been trying is this:

    int c;
     char *w, test;
     w=&test;
    printf("Give product code: ");
    while (c = getchar() != EOF)
        {
        *w=c;
        w=w+1;
        }
      w=w+1;
       *w='\0';
 
 
    printf("%s", *w);

However, I always seem to get (null) as the printf result. I even tryed setting a counter and printing the *(w-counter) result but yet again... no luck. Any help really appreciated. :)

(By the way, on a side note. Is there any way I can avoid the declaration of "test" and "w=&test"?)

Thank you very much. :)

[Savage]

Hello there.

I'm (kind of) new to programming as you can see, and I have a question.

I want the user to input a string (for example: a product code) but store it dynamically. There must be an other way instead of a static array with, let's say, 50 elements. What I have been trying is this:

    int c;
     char *w, test;
     w=&test;
    printf("Give product code: ");
    while (c = getchar() != EOF)
        {
        *w=c;
        w=w+1;
        }
      w=w+1;
       *w='\0';
 
 
    printf("%s", *w);

However, I always seem to get (null) as the printf result. I even tryed setting a counter and printing the *(w-counter) result but yet again... no luck. Any help really appreciated. :)

(By the way, on a side note. Is there any way I can avoid the declaration of "test" and "w=&test"?)

Thank you very much. :)

And how did u exectly used the counter?

Savage

[Avon]

Well I added the counter inside the "while" loop but I didn't have high hopes anyway. ;) Here is what it looked like.

        int c,counter=0;
     char *w, test;
     w=&test;
    printf("Give product code: ");
    while (c = getchar() != EOF)
        {
        *w=c;
        w=w+1;
                counter=counter+1;
        }
      w=w+1;
       *w='\0';
 
 
    printf("%s", *(w-counter));

[Savage]

Well I added the counter inside the "while" loop but I didn't have high hopes anyway. ;) Here is what it looked like.

        int c,counter=0;
     char *w, test;
     w=&test;
    printf("Give product code: ");
    while (c = getchar() != EOF)
        {
        *w=c;
        w=w+1;
                counter=counter+1;
        }
      w=w+1;
       *w='\0';
 
 
    printf("%s", *(w-counter));

But,have u tryed this:

    while (c = getchar() != EOF)
        {
           *(w+counter)=c;
                                   counter=counter+1;
                                }

???

Savage

[weaknessforcats]

THis code:

int c,counter=0;
     char *w, test;
     w=&test;
    printf("Give product code: ");
    while (c = getchar() != EOF)
        {
        *w=c;
        w=w+1;
                counter=counter+1;
        }
      w=w+1;
       *w='\0';
 
 
    printf("%s", *(w-counter));
 

Has some problems. First, test is 1 char. w is a pointer to that char. That means you can w=w+1 becuse you don't own the location w+1.

You will need to build an array. Check out strcat().

[Avon]

THis code:

int c,counter=0;
     char *w, test;
     w=&test;
    printf("Give product code: ");
    while (c = getchar() != EOF)
        {
        *w=c;
        w=w+1;
                counter=counter+1;
        }
      w=w+1;
       *w='\0';
 
 
    printf("%s", *(w-counter));
 

Has some problems. First, test is 1 char. w is a pointer to that char. That means you can w=w+1 becuse you don't own the location w+1.

You will need to build an array. Check out strcat().

Ok thanks for the reply. I understand what the problem is, however, I don't unterstand how to use the array. Am I going to use it in a way that it stores each separate string until you call the function again, or store all strings? If you mean the second, how do we know the size of the array? :)

[weaknessforcats]

To store an array of 10 strings you need an array of char*. Either:

char* arr[10];
 

or

char** arr = new char*[10];
 

These two declaratiions are equivalent. The only difference is that you have to delete the array you allocate with new.

So, now you have an array of char*.

Get your string into a buffer.

Now just allocate memory of the right amount and attach it to one of your char* array elements. Then copy the string to your memory allocation:

arr[i] = new char[strlen(buffer) + 1];  //+1 for the null terminator
strcpy(arr[i], buffer;
 

and off you go.