Thread: An uninitialized array value = 4198445?

I'm attempting to learn C at my school, and I'm going through Stephen Kochan's Programming in C book. I just recently tried to compile 7.1, but I got some definately odd results.

Here's the code itself:

Code:

#include <stdio.h>


int main (void)
{
 int values[10];
 int index;

 values[0] = 197;
 values[2] = -100;
 values[5] = 350;
 values[3] = values[0] + values[5];
 values[9] = values [5] / 10 ;
 --values[2];

 for ( index = 0; index < 10; ++index )
 printf ("values[%i] = %i\n", index, values[index]);

 return 0;
 }

And my output:

Code:

values[0] = 197
values[1] = 4198445
values[2] = -101
values[3] = 547
values[4] = 37814120
values[5] = 350
values[6] = 37814124
values[7] = 895
values[8] = 37814176
values[9] = 35

Shouldn't the uninitialized values (1, 4 and 6-8) be zero? Could somebody explain this to me please?

I believe if an array element is not explicitely initialized it can be anything, typically what was already in that memory address.

Shouldn't the uninitialized values (1, 4 and 6-8) be zero?

no! they should be 'undefinded'--which it looks like they are.....

in other words, if you don't initialize them, they equal whatever is stored in that memory location. in other other words, garbage.

initialize your array, then do your work. you'll find that better in the long run.

Oh, okay. The book doesn't make a big point of explaining what was going on there. Thanks for letting me know what was going on.

but do realize that

double a[5];

and

double *a = malloc(5 * sizeof *a);

is 2 completely different things. The first gets destroyed automatically when it goes out of scope, the second has to be freed somewhere.

Um.... 'Kay. A lot of this is going to be helpful in the future, I'm assuming? A lot of this is out of my knowledge thus far....