Why does this initialize the first element only? - c

Question

I'm trying to figure out why initializing an array with int buckets[AS] = { 0 }; is not setting all the elements to zero. Perhaps it is a compiler optimization, in which case would volatile be acceptable? volatile int buckets[AS] = { 0 };.
Second question, why is only the first element initialized to 1 here? Doesn't this fall under:
C99 [$6.7.8/21]
If there are fewer initializers in a brace-enclosed list than there are elements or members
of an aggregate, or fewer characters in a string literal used to initialize an array of known
size than there are elements in the array, the remainder of the aggregate shall be initialized implicitly the same as objects that have static storage duration.
Problem code:
#include <stdio.h>
#define AS 100
int buckets[AS] = { 1 };
int main()
{
int i;
for(i = 0; i < AS; i++) {
printf("%d", buckets[i]);
}
return 0;
}
EDIT:
Changing optimization level from -o0 to default eliminates the issue. Working with STM32 Kiel IDE micro-controllers.
EDIT EDIT:
This is the code causing trouble. Compiler optimizing away for loop?
// Initialize this to 1 as when initializing to 0 and clearing some elements are non-zero
// Possibly a compiler bug? Changing optimization level from -o0 to default eliminates the issue
uint16_t pulse_time_hist[NUM_BUCKETS] = {1};
// Resets all values stored in the histogram
void clearHist() {
unsigned int i;
for (i = 0; i < NUM_BUCKETS; i++) {
pulse_time_hist[i] = 0;
}
}
EDIT EDIT EDIT:
I'm not a compiler guy at all btw. Here is my compiler control string
-c -cpu Cortex-M4.fp -D__EVAL -g -O0 -apcs=interwork -split_sections ...
Running without -c99 currently.

Answer 1

In the quotation "... shall be initialized implicitly ..." a value for initialization is not mentioned, so rest of array is initialized by default value, and for number types this default value is 0.
To continue the topic try the following snippet
#include <stdio.h>
#define N 10
int arr1[N];
int main()
{
int arr2[N];
int i;
for (i = 0; i < N; i++) {
printf("%d ", arr1[i]);
}
printf("\n");
for (i = 0; i < N; i++) {
printf("%d ", arr2[i]);
}
return 0;
}
and try to understand why the output is
i.e. why global/static objects (like arr1) are initialized without initializers, but local/automatic/stack-allocated (like arr2) are not.
UPDATE:
Section 6.7.8 Initialization of C99 standard said:
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.

Answer 2

This is because the missing values are automatically initialized to zero. If you want to initialize every element to 1, you need to have this :
int buckets[AS] = {1, 1, 1, ..., 1}; //100 times
which clearly isn't feasible.
You may also want to read this answer as well.

Answer 3

int buckets[AS] = { 1 };
In which case you can omit some part of the initializer and the corresponding elements will be initialized to 0.
int buckets[AS] = { 1 }; //initialize to 1,0,0,0,0...
If you are trying to initialize all elements of the array to 1 there is a GCC extension you can use with the following syntax:
int buckets[AS] = {[0 … 99] = 1 };