### Tip 41: Initialize arrays and structs with zeros

level: struct user
purpose: don't leave values undefined

Part of a series of tips on POSIX and C. Start from the tip intro page, or get 21st Century C, the book based on this series.

If you declare a variable inside a function, then C won't zero it out automatically. Which is odd, because these are known as automatic variables. For more on the memory types, see Tip #20. I'm guessing that the rationale here is a speed savings: when setting up the frame for a function, zeroing out bits is extra time spent, which could potentially add up if you call the function a million times and it's 1985.

But here in the present, leaving a variable undefined is asking for trouble.

For simple numeric data, set it to zero on the line where you declare the variable. For pointers, including strings, set it to NULL. OK, that's easy enough, as long as you remember.

For structs and arrays of constant size, recall how last time I showed you that if you use designated initializers but leave some elements blank, those blank elements get set to zero; you can set the whole structure to zero by assigning a complete blank. Here's a do-nothing program to demonstrate the idea:

typedef struct {
int la, de, da;

int main(){
int ll[20] = {};
}


Isn't that easy and sweet.

Now for the sad part: let us say that you have a variable-length array (i.e., one whose length is set by a run-time variable). The only way to zero it out is via memset:

int main(){
int length=20;
int ll[length];
memset(ll, 0, 20*sizeof(int));
}


This is bitter in exact proportion to the sweetness of initializing fixed-length arrays. You can blame ISO C standard part 6.7.8§3 for this, because it insists that variable length arrays can't be initialized. I say the compiler should be able to work it out.... So it goes.

To do:
Write yourself a macro to declare a variable-length array and set all of its elements to zero. You'll need inputs listing the type, name, and size.