Variable initialization evaluates to something unexpected

Question

Why does the int variable nl evaluate to 33881 if you call a printf("%d",nl); after this the following line?

nl, nw, nc = 0;

How could it possibly evaluate to any other value than 0 before the loop? I am compiling with gcc from the terminal in a crouton/Ubuntu 12.04LTS environment on my ARM Chromebook, so I don't know if this is a bug in the code or my machine.

#include <stdio.h>

#define IN 1
#define OUT 0

main() {
  int c, nl, nw, nc, state;
  state = OUT;
  nl, nw, nc = 0;
  while ((c = getchar()) != EOF ) {
    ++nc;
    if (c == '\n') {
      ++nl;
    }
    if (c == ' ' || c== '\n' || c== '\t') {
      state = OUT;
    } else if (state == OUT) {
      state = IN;
      ++nw;
    }
  }
  printf("%d %d %d\n", nl,nw,nc);
}

I tried adding another declarative line nl = 0; to force the program to return predictable values, but I still am no closer to understanding this behavior.

OT: It's ìnt main(void) at least.

alk
– alk

2015-02-22 16:57:31 +00:00
Commented Feb 22, 2015 at 16:57 — alk
– alk, Commented Feb 22, 2015 at 16:57

Jeffrey Bosboom · Accepted Answer · 2015-03-14 02:07:21Z

1

This

nl, nw, nc = 0;

does not initialize nl; if you enable compiler warnings it will tell you that the left operand of the comma operator has no effect, because the , operator ignores all the left operands and keeps the last one. To initialize all the variables at once you need this:

nl = nw = nc = 0;

Since nl is uninitialized, it contains a garbage value; you need to explicitly initialize it.

Compiler warnings would also complain about nl and nw being uninitialized.

edited Mar 14, 2015 at 2:07

Jeffrey Bosboom

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answered Feb 22, 2015 at 16:51

Iharob Al Asimi

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2 Comments

user134143 Over a year ago

So the evaluation is the "standard" or "placeholder" value assigned to any variable of type int that just hasn't been declared as something else, Thank you. I just feel silly now.

Iharob Al Asimi Over a year ago

@user134143 no, variables are not initialized when you declare them, it's just that the memory location where they are stored contains something that then is overwritten when you initilize it, it's done that way for performance reasons.

Iharob Al Asimi · Accepted Answer · 2015-02-22 17:01:44Z

0

The line

nl, nw, nc = 0;

will initialize the variable "nc", but the variables "nl" and "nw" will receive some random number.

To initialize all three you must use

nl = nw = nc = 0;

edited Feb 22, 2015 at 17:01

Iharob Al Asimi

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answered Feb 22, 2015 at 17:00

Cleton

11 bronze badge

3 Comments

user134143 Over a year ago

Would it be a coincidence (as in the memory of where it was allocated was holding "0")if I told you that the nw variable evaluated to 0 as well, only the nl was throwing that strange number,

Iharob Al Asimi Over a year ago

@user134143 it's nothing but a coincidence.

Cleton Over a year ago

I think so, because when your program asks the OS to allocate the memory for a variable, for example an int, the OS reserve 16 bits (on most systems) for that variable and these 16 bits can contain any value.

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