Define integer (int); What's the default value?

Question

int i;
int data[5] = {0};
data[0] = i;

What's the value in data[0]?

Also, what's the meaning of this line?

if (!data[0]) { ... }

Answer 1

In most cases, there is no "default" value for an int object.

If you declare int i; as a (non-static) local variable inside of a function, it has an indeterminate value. It is uninitialized and you can't use it until you write a valid value to it.

It's a good habit to get into to explicitly initialize any object when you declare it.

Answer 2

It depends on where the code is written. Consider:

int i;
int data[5] = {0};

void func1(void)
{
    data[0] = i;
}

void func2(void)
{
    int i;
    int data[5] = {0};
    data[0] = i;
    ...
}

The value assigned to data[0] in func1() is completely deterministic; it must be zero (assuming no other assignments have interfered with the values of the global variables i and data).

By contrast, the value set in func2() is completely indeterminate; you cannot reliably state what value will be assigned to data[0] because no value has been reliably assigned to i in the function. It will likely be a value that was on the stack from some previous function call, but that depends on both the compiler and the program and is not even 'implementation defined'; it is pure undefined behaviour.

You also ask "What is the meaning of this?"

if (!data[0]) { ... }

The '!' operator does a logical inversion of the value it is applied to: it maps zero to one, and maps any non-zero value to zero. The overall condition evaluates to true if the expression evaluates to a non-zero value. So, if data[0] is 0, !data[0] maps to 1 and the code in the block is executed; if data[0] is not 0, !data[0] maps to 0 and the code in the block is not executed.

It is a commonly used idiom because it is more succinct than the alternative:

if (data[0] == 0) { ... }

Answer 3

if an integer is declared globally then it is initialized automatically with zero but if it is local then contains garbage value until and unless itis given some value

Answer 4

If an integer is not initialized, its value is undefined as per C

Answer 5

Since you've included the ={0};, the entire array is filled with zeros. If this is defined outside any function, it would be initialized with zeros even without the initializer. if (!data[x]) is equivalent to if (data[x] == 0).

Answer 6

// File 'a.c'

   #include <stdio.h> 

   void main()
    {
            int i, j , k;
            printf("i = %i j = %i k = %i\n", i, j, k);
    }

// test results

> $ gcc a.c 
> $ ./a.out 
> i = 32767 j = 0 k = 0

Answer 7

As @James McNellis stated in 2010, the standards for programming languages leave the initial value of a variable to the compiler implementation, and results differ from one compiler implementation to another.

I came here to review the state of affairs in this regard because I remembered observing that some compilers initialize arrays of integers to zero.

After reading the inconclusive answers recorded here, I constructed a small static method in the assembly that I use to conduct such experiments before I implement a technique in production code. My current project requires porting some JavaScript code to C#. Since the JavaScript interpreter doesn't initialize anything, that implementation opened with a FOR loop to initialize a large array of integers. Before I blindly left it in the C# implementation.

The method follows.

        private static void IntegerArrayGym ( )
    {
        int [ ] ints = new int [ 10 ];

        Console.WriteLine ( $"{TEST_INTEGER_ARRAY_GYM} Begin:{Environment.NewLine}" );

        int intArrayLength = ints.Length;

        Console.WriteLine ( $"    The test array consists of {ints} uninitialized integers.{Environment.NewLine}" );

        for ( int intJ = ArrayInfo.ARRAY_FIRST_ELEMENT;
                  intJ < intArrayLength;
                  intJ++ )
        {
            Console.WriteLine ( $"    Value of integer at ordinal position {ArrayInfo.OrdinalFromIndex ( intJ )} = {ints [ intJ ]}" );
        }   // for ( int intJ = ArrayInfo.ARRAY_FIRST_ELEMENT; intJ < intArrayLength; intJ++ )

        Console.WriteLine ( $"{Environment.NewLine}{TEST_INTEGER_ARRAY_GYM} Done!" );
    }   // private static void IntegerArrayGym

The output follows.

CSharp_Lab, version 7.92.170.0 Begin @ 03/06/2023 22:58:49.124 (03/07/2023 04:58:49.124 UTC)

IntegerArrayGym Begin:

The test array consists of System.Int32[] uninitialized integers.

Value of integer at ordinal position 1 = 0
Value of integer at ordinal position 2 = 0
Value of integer at ordinal position 3 = 0
Value of integer at ordinal position 4 = 0
Value of integer at ordinal position 5 = 0
Value of integer at ordinal position 6 = 0
Value of integer at ordinal position 7 = 0
Value of integer at ordinal position 8 = 0
Value of integer at ordinal position 9 = 0
Value of integer at ordinal position 10 = 0

IntegerArrayGym Done! Please press the ENTER (Return) key to exit the program.