Default Char Value - Simple Question

[jmpeer]

Yo, look at this simple code:

#include <stdio.h>

int main()
{
    unsigned char bytes[2][100];
    int i = 0;
    for(i = 0; i < 200; i++)
    {
        printf("Byte[0][%-3d]: %-5d \t Byte[1][%-3d]: %-5d \n",i,bytes[0][i],i,bytes[1][i]);
    }
    return 0;
}

In C, does the datatype char have a default value?
This gives me a random series of values.
And even though the array only goes to 100, it prints up to 200 like I specified without error... and more random values.
Why is that?
Where are these numbers coming from?

[laserlight]

In C, does the datatype char have a default value?

Not really. You might consider the null character as a default, but then there is no "default construction".

This gives me a random series of values.
And even though the array only goes to 100, it prints up to 200 like I specified without error... and more random values.
Why is that?

Basically, your array is left uninitialised.

Where are these numbers coming from?

Presumably whatever was present in memory.

To answer your unasked question, you probably want to initialise, e.g.,

unsigned char bytes[2][100] = {""};

[slingerland3g]

What you did was request that the compiler set up two blocks of memory each 100 chars in size. In your for loop you pretty much did a common mistake of overstepping your array bounds. Since you did not initialize your two chunks of memory, what the compiler gave you was two blocks of available memory that has not be locked or in use by another process. What ever is there was what has been given up by a prior process called from free() or what ever other memory utility that will free memory up.

[jmpeer]

Wow. C doesn't really take care of anything for you.

Alright.

That's all I need to know.

[nonoob]

That's why it's fast. There is nothing "done for you" to waste time when you didn't need it done.