Thread: and now pointers are pointing in every direction

OK so here it is:

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

main()
{
char names[20] = "\0";
char * pNames = names;
char * pList[3] = {"\0", "\0", "\0"};
int i = 0;

for (i = 0; i < 3; i++)
{printf("Enter a name: \n");
scanf("%19s", pNames);
pList[i] = pNames;}
for (i = 0; i < 3; i++)
{printf("%s", pList[i]);}
return 0;
}

it outputs the last name entered 3 times. What am I doing wrong. the intention here is to input 3 names and then output them one after the other.

Thanks in advance for putting up with my noobish nature.

I don't really understand what you are trying to do with pNames and pList, but this should do what you want:

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

int main()
{
char names[3][20];
int i = 0;

for (i = 0; i < 3; i++)
{
printf("Enter a name: \n");
scanf("%19s", names[i]);
}
for (i = 0; i < 3; i++)
{
printf("%s\n", names[i]);
}
return 0;
}

Why would you want 2 pointers (names, pNames) pointing to the same memory to achieve this?

char names[3][20] so this is the trick. the first notation is number of strings and secone is string length. the way my book describes it is waaaaay different lol.

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I just realized that this does not use a pointer. I'm trying to get familiar with them as they are supposed to act more like strings. at least acording to my book they do.

Originally Posted by Mattfodder

I just realized that this does not use a pointer. I'm trying to get familiar with them as they are supposed to act more like strings. at least acording to my book they do.

If you write "char arr[30]" then arr is a pointer. You could also write "char *ptr", but then you would have to allocate and free the memory you need yourself. With the first method it is done for you.
Arrays are Pointers to the memory address where the data is stored in C.

so if you are going to use a pointer array you need to use malloc?

Originally Posted by Mattfodder

so if you are going to use a pointer array you need to use malloc?

Yeah the memory at a pointers address isn't allocated for you.
The exception is of course, when you declare arrays like above, but then you only use pointers implicit.