malloc

char *pointer[100];
.
.
.
int input()
{
int i=0; int length; char help[100];

printf("Input: ");
scanf("%s",help);
while((strcmp(help,"ENDE")!=0)&&(i<100))
{
length=strlen(help);
pointer[i]=(char*)malloc((length+1)*sizeof(char));
pointer[i]=help;
i++;
printf("Input: ");
scanf("%s",help);
}
return i;
}

if I enter a word and assign it to pointer[i] (how showed above), the program overwrites all elements (pointer[0] until pointer[i-1]) with this last entered word. if I want to output all elements at the end of the program (not showed above) the program puts out strange signs (always the same "smiley").

Whats going wrong here??????

Code:

char *pointer[100];
...
pointer[i]=(char*)malloc((length+1)*sizeof(char));

Don't do that! It does not make any sense. You should use an array of structures to save the strings.

Code:

struct{
    char *pointer;
}mystruct[100];
...
length=strlen(help);
mystruct[i].pointer = malloc((length)*sizeof(char));
...
for(k=0; k < length; k++)
{
    mystruct[i].pointer[k] = help[k];
}

[edit]
Hmm .. I've just checked my suggested source code and found that it doesn't work for certain string sizes (strange output). Maybe you should use char arrays of fixed size in the structure. That should fix the problems.

Code:

struct{
    char pointer[100];
}mystruct[100];
...
length=strlen(help);
...
for(k=0; k < length; k++)
{
    mystruct[i].pointer[k] = help[k];
}

[/edit]

And better learn how to use fgets() instead of scanf(). It will save you lots of trouble. Never forget: fgets() does not remove the '\n' at the end of a line!

You should use a for loop:

Code:

for(x=0;x<100;x++)
{
    printf("Input: ");
    scanf("%s",help);
    if(strcmp(help,"ENDE")==0)
    {
        break; //leave the for loop
    }
    length=strlen(help);
    ....
}

First, declaring an array of pointers to char is a good way. But that is not the problem here, the problem is this:

pointer[i]=help;

To copy strings, you should use the function strcpy (string copy). For a definition of strcpy and a quick reference to C standard functions, look here:
http://www.acm.uiuc.edu/webmonkeys/b...14.html#strcpy

Casting of malloc isn't also necessary in ANSI C, just include stdlib.h.

Originally posted by Shiro
First, declaring an array of pointers to char is a good way

Yes, of course it is okay! I should have pointed out, it was the conjunction with malloc() that made me think:

pointer[i]=(char*)malloc((length+1)*sizeof(char));

I don't see how this would do any good. As far as I understand Sylli22 wants to store pointers to arrays (each containing an input string) in the array of pointers char *pointer[100].
But the line of source code above won't do the job.
Please, correct me, if I am wrong!

As I understood the code of Sylli22, he/she wants to fill the array pointer with 100 or less strings. The variable pointer is an array of 100 pointers to char.

The mentioned line allocates space for (length+1) chars, this space is pointed to by a pointer to char. Since pointer[i] is the i-th pointer to char in the array, this is correct.

Note that Sylli22 is creating a 2D array here where each row has a variable length, which is the string length of the string where each element of the array points to. So pointer[i] points to the i-th string in the 2D array.

Ah! Now I got it! Thank you for your explanation, Shiro.
I was just wondering "How the hell would you access the allocated space?" .. I was thinking 1D not 2D.

thank you, now, it works very good. the mistake was the missing strcpy(); thank you very much.