未初始化为0时的有趣值

Question

I have the following code which is supposed to display each input word on a separate line. For kicks, I placed a printf to display the word count after every word to make sure I understood the logic.

#include <stdio.h>

/*Write a program that prints its input one word per line*/

#define IN 1
#define OUT 0

int main()
{
    int c, state, wordcount;

    state = OUT;
    while ((c = getchar()) != EOF) 
    {
        if (c == ' ' || c == '\t' || c == '\n') 
        {
            if (state == IN)
            {
                printf("\n");
                printf("Current word count: %i\n", wordcount); 
                /*troubleshooting*/
                state = OUT;
            }
        }
        else if (state == OUT) 
        {
            state = IN;
            putchar(c);
            ++wordcount;
        }
        else
        {
            putchar(c);
        }
    }
}

The output for this is shown in this picture. For some reason, without initializing the wordcount variable initially to 0, it starts at 9.

Wordcount variable gets set to 9 when it is not initialized to 0

When initializing the wordcount variable to 0 (just setting wordcount = 0 in the int declaration statement), everything works as expected:

Wordcount variable is correct when initialized to 0

Can someone explain to me what is going on here? Does this have something to do with how these variables are stored in memory? Trying to understand what is going on. Thanks!

Answer 1

Before the OP edited the question:

This is Undefined Behavior.

if (state == IN)
//Some code
else if (state == OUT)
//Some code

It looks like State is not initialized (or set to a value) before comparing it. Leaving a value uninitialized is not Undefined Behavior, but accessing this way is definitely way to get in horrors of UB!

In C89 , see section 6.5.7 Initialization .

If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate. If an object that has static storage duration is not initialized explicitly, it is initialized implicitly as if every member that has arithmetic type were assigned 0 and every member that has pointer type were assigned a null pointer constant

In C99 , see section 6.7.8 Initialization

If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate. If an object that has static storage duration is not initialized explicitly, then: — if it has pointer type, it is initialized to a null pointer; — if it has arithmetic type, it is initialized to (positive or unsigned) zero; — if it is an aggregate, every member is initialized (recursively) according to these rules; — if it is a union, the first named member is initialized (recursively) according to these rules.

Refer standard section 6.3.2.1p2 :

If the lvalue designates an object of automatic storage duration that could have been declared with the register storage class (never had its address taken), and that object is uninitialized (not declared with an initializer and no assignment to it has been performed prior to use), the behavior is undefined.

According to C99 Standard, its "indeterminate value", it is either an unspecified value or a trap representation.

But, before you get into the minors details these standards provide, you should acquire some of the most common and good-to-know coding habits. This can be a good start.

---

After the edit from OP:

I ran the exact same code from OP which has state = OUT; ie State initialized, it behaves the way it should.

stdin:

Lazy Frog

Output:

Lazy
Current word count: 1
Frog
Current word count: 2

Answer 2

If you initialize a variable, it gets a "random" (unused) allocation in your RAM, if something has been already written there, the variable uses this value for itself.

That's why you always declare variables.

The backend is a bit more complicated, but thats the "simple" explanation.