C Programming Help (Pointers/Arrays)

[RZetlin]

This program is suppose to get the inputs and place the inputs into a pointer array.

Eg.
INPUT:
Hello
Bye
OUTPUT:
Hello
Bye

This is what is occurring:
INPUT:
Hello
Bye
OUTPUT:
Bye
Bye

Can someone please help me!

#include stdio.h (Can't show <> for some reason)

int main()
{
char line[20];
char *list[50];
int counter=0,count;

gets(line);
list[counter]=line;
counter=counter+1;
gets(line);
list[counter]=line;

for (count=0;count<=counter;count++)
{
puts (list[count]);
}

return 0;
}

[Mohd]

do the following because when you change line in the second time the pointer list[counter] is point to line and line is changed

int main()
{
char line[20];
char line2[20];
char *list[50];
int counter=0,count;

gets(line);
list[counter]=line;
counter=counter+1;
gets(line2);
list[counter]=line2;

for (count=0;count<=counter;count++)
{
puts (*(list+count));
}

return 0;
}

[RZetlin]

Originally posted by squall84
do the following because when you change line in the second time the pointer list[counter] is point to line and line is changed

Hmm...that won't work if the input is in a for loop.

#include stdio.h (Can't show <> for some reason)

int main()
{
char line[20];
char *list[50];
int counter=0,count;

for (counter=0;counter<2;counter++)
{
printf("Enter Line: %d",count);
gets(line);
list[counter]=line;
}

for (count=0;count<=counter;count++)
{
puts (list[count]);
}

return 0;
}

Anyway to work around that?

[ammoQ]

Well, a pointer array is just an array of pointers. But you need something else: Space to store your data. A pointer is simply an address, not the space where you store your data. Easiest solution: Use a string array instead of a pointer array.

#include &lt;stdio.h&gt; // (Can show &lt;&gt; for a good reason ;-)

int main()
{
char list[50][20];
int counter=0,count;

counter=0;
// you would normaly place some kind of loop here
gets(list[counter++]);
gets(list[counter++]);
// end loop

for (count=0;count&lt;counter;count++)
{
puts (list[count]);
}

return 0;
}

[ammoQ]

Now the solution with pointers:

#include &lt;stdio.h&gt; // (Can show <> for a good reason ;-)
#include &lt;stdlib.h&gt;

int main()
{
char *list[50];
int counter=0,count;

counter=0;
// you would normaly place some kind of loop here
list[counter] = malloc(20); // get some memory
gets(list[counter++]);
list[counter] = malloc(20);
gets(list[counter++]);
// end loop

for (count=0;count&lt;counter;count++)
{
puts (list[count]);
free(list[count]); // give back the memory
}

return 0;
}

[Mohd]

Thanks ammoQ, I was looking for another use of function malloc (memory allocation).

by the way in my study we didn't discuss function malloc yet so if this is my assingment my teacher will not accept it with malloc function.

[ammoQ]

Without malloc, you have to pre-allocate memory in an array, like in my first example. Note: all those examples are totaly unsafe in terms of buffer overflows etc. There is no way to use gets() safely, it would be better to use fgets(s, 20, stdin) instead.