diffculty in entering strings in C using pointer

Question

The following program works in gcc but on giving the value of T = 6, this program continues and does not end on asking for input strings. Any help guys if you recognise whats wrong with this program?

int main()
{ int T,i,j;
  char *strings[T];
  printf("Enter the Number of Strings to Reverse : \n");

  scanf("%d ",&T);

  for(i=0;i<T;i++)
   { strings[i] = (char *)malloc(100*sizeof(char));
     scanf("%s\n",strings[i]);
   }
  for(i=0;i<T;i++)
   {printf(" The String %d is : %s\n",i+1,strings[i]);
  }
return 0;
}

`T` is uninitialized when you are creating your pointer array `strings` — Andreas Fester, Mar 05 '14 at 18:45
@Andreas got it, but the array is 0 length (T would be initialized to 0), right? — John Yost, Mar 05 '14 at 18:48
But then `strings` is already allocated. Read http://stackoverflow.com/questions/14049777/why-global-variables-are-always-initialized-to-0-but-not-local-variables — Andreas Fester, Mar 05 '14 at 18:52
@user1463736 but at that moment `strings` will be initialized already with `0` — Dabo, Mar 05 '14 at 18:52
@JohnYost local variables cannot be counted on to be initialized implicitly, so no, T would not (reliably) be initialized to 0. — mah, Mar 05 '14 at 18:52
@Dabo if i input the value of T as 1 or 2 or 3 till 5 answer comes but whats wrong with it with T = 6. Please run it once on gcc and let me know the error — user1463736, Mar 05 '14 at 18:58
@Andreas if i input the value of T as 1 or 2 or 3 till 5 answer comes but whats wrong with it with T = 6. Please run it once on gcc and let me know the error — user1463736, Mar 05 '14 at 19:01

score 2 · Answer 1 · answered May 20 '17 at 20:08

2

T is not initialised (Remember in C++, local scope variables are not automatically initialised) :

int T= 6;

answered May 20 '17 at 20:08

Jeshen Appanna

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score 1 · Accepted Answer · answered Mar 05 '14 at 19:11

1

T is not initialized inside main() therefore has an undefined value.

char *strings[T] creates an array of char * pointers of an undefined length.

Fix this using:

int T=6;

Or, given T is in fact constant:

const int T=6

or perhaps better

#define T 6

Feel free to use a more mnemonic name than T.

answered Mar 05 '14 at 19:11

abligh

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but T value will be given by user when program executes. – user1463736 Mar 05 '14 at 19:20
@user1463736, yes, but then the array of pointers is **already allocated** in memory! – Stefan Marinov Mar 05 '14 at 19:21
@smarinov ok got it now. Thanks everyone for the help ! – user1463736 Mar 05 '14 at 19:25
1

You could move the declaration of `strings` after `T` has been read. However, you are relying on (I believe) a gnu extension for that to work. A more common way would be to dynamically allocate a data structure of size `T * sizeof (char *)` using `malloc()`. – abligh Mar 05 '14 at 19:27

diffculty in entering strings in C using pointer

2 Answers2