Thread: array of pointers passed as argument to fgets

Hi Everyone!
I am trying to use array of pointers to store strings and trying to take the input from the user for that I am using fgets,but it's throwing a segmentation fault.what is the right syntax when dealing with array of pointer with fgets?
Program :
Trying to store 3 strings:name,email id and subject,for this I am using char array of pointers.

Code:

#include<stdio.h>
int main()
{
    char *info[3];
    fputs("enter name:",stdout);
    fgets(info[0],sizeof(info),stdin);  // this line is throwing segmentation fault
    fputs(info,stdout); 
    return 0;
}

Originally Posted by Salem

Well you need to allocate space for the strings.

1.
info[0] = malloc( 100 );
fgets(info[0],100,stdin);


2.
char info[3][100];
fgets(info[0],sizeof(info[0]),stdin);

Code:

#include<stdio.h>
int main()
{
    char *days[2]={"mon","tue"};
    printf("%s\n",days[0]);
     return 0;
}

Where in data mon & tue getting stored?
Instead of initialising in the program I want give input from the keyboard.

I am tryin to use char array of pointer instead of a 2d char array,to reduce the fixed size usage of the 2d array.

Originally Posted by shruthi

I am tryin to use char array of pointer instead of a 2d char array,to reduce the fixed size usage of the 2d array.

If each word has three letters, there would be no space saving, since either type of array entry, would need 3 letters + 1 end of string char, for each word in the array.

For example:

Code:

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

int main(void) {
   char *day[7];
   int i;
   for(i=0;i<7;i++) {
      day[i]=malloc(4);
      printf("Enter a three letter day abbreviation: ");
      
      fgets(day[i], 4,stdin);
      getchar();
      printf("%d. day: %s\n",i+1,day[i]);
   }
   for(i=0;i<7;i++)
      printf("%d) day: %s\n",i+1,day[i]);
   printf("\n");
   return 0;
}

To save space, you'd want to put the user's entry into a second array, and then measure it's length (and probably remove the \n which fgets() adds to the string).

Once you have the length of the entry, you could use that variable in malloc(), instead of a constant like here (the 4). Then copy the string into the final position in the array of pointers to strings.

Give that a try.

Adak - You forgot to free the memory you allocated.

Originally Posted by Click_here

Adak - You forgot to free the memory you allocated.

Two reasons:

#1) I don't believe this is the type of program that he wants, since there is no space being saved, which is what he wants to do.
This is "some" code, based on his example, but not "the answer" to his question. I intended that the OP would post back, asking "OK, how can I do this then, so I can save space?"

#2) I'm running in a virtual environment. All memory is freed, anyway. For small programs that I will probably never run again, I don't always bother with it, anymore. Good on ya, if you still free all the pointers, for every little snippet of code.

Adak:
"For small programs that I will probably never run again, I don't always bother with it"

I didn't know we could do that

So just to clarify - You do not need to free memory that you allocate using malloc, because it will automatically be returned when the process is terminated.

I thought that it was a memory leak if you didn't free the allocated memory - I'm always happy to learn something new

Originally Posted by Adak

If each word has three letters, there would be no space saving, since either type of array entry, would need 3 letters + 1 end of string char, for each word in the array.

For example:

Code:

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

int main(void) {
   char *day[7];
   int i;
   for(i=0;i<7;i++) {
      day[i]=malloc(4);
      printf("Enter a three letter day abbreviation: ");
      
      fgets(day[i], 4,stdin);
      getchar();
      printf("%d. day: %s\n",i+1,day[i]);
   }
   for(i=0;i<7;i++)
      printf("%d) day: %s\n",i+1,day[i]);
   printf("\n");
   return 0;
}

To save space, you'd want to put the user's entry into a second array, and then measure it's length (and probably remove the \n which fgets() adds to the string).

Once you have the length of the entry, you could use that variable in malloc(), instead of a constant like here (the 4). Then copy the string into the final position in the array of pointers to strings.

Give that a try.

I guess it's just easier to use a 2d char array.
I thought since you can store strings

Code:

char *arr[3]={"mon","tuesday","wednesday"};

I thought one could possibly take the user input and it would get automatically stored the same way how I have show it above.
I was just trying to use that advantage of array of pointers to 2d char array.But what I have understood is you still have to create memory,I think I'll just use 2d char array instead of calculating the length and again putting it in to the array.

Originally Posted by Adak

@Click_here: For a program of any importance, yes, you should free the memory you allocate.

@shruthi: You CAN save memory by using 2D (pointer to pointer) arrays, just like you were thinking about. It's just not "automatic", like the pre-defined strings you show: "mon", "tuesday", "wednesday". Each string needs to have it's length known, and the memory allocated for it. This would be a better program, (and a useful example), so it would be VERY worthwhile to free any allocated memory.


If you are ready to study and or practice using char pointers, it's a very worthwhile skill to learn how it works.

Ok,Ive given it a try.

Code:

 #include#include<stdio.h>
#include<string.h>
#include<stdlib.h>
 int main()
{
char *str1[3];
char data[40];
int len;
fgets(data,sizeof(data),stdin);
fputs(data,stdout);
len=strlen(data);printf("length of the string=%d\n",len);
str1[0]=malloc(strlen(data));
strncpy(str1[0],data,len);printf("%s\n",str1[0]);
free(str1[0]);
return 0;
}

Able to take the input,find out it's length,create a memory of that size & copy the data into that,but the only thing is not able to place the null character after copying.