How to Initialize a Struct to 0 in C++

Initializing a struct to 0

The first is easiest(involves less typing), and it is guaranteed to work, all members will be set to 0^{[Ref 1]}.

The second is more readable.

The choice depends on user preference or the one which your coding standard mandates.

^{[Ref 1]} Reference C99 Standard 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.

Good Read:
C and C++ : Partial initialization of automatic structure

How to initialize a struct to 0 in C++

Before we start:

1. Let me point out that a lot of the confusion around this syntax comes because in C and C++ you can use the = {0} syntax to initialize all members of a C-style array to zero! See here: https://en.cppreference.com/w/c/language/array_initialization. This works:

   // z has type int[3] and holds all zeroes, as: `{0, 0, 0}`
   int z[3] = {0}; 
   

   But, that syntax does not work the same for structs, which are entirely different animals than C-style arrays.

2. See also my follow-up question I asked after writing this answer below: Why doesn't initializing a C++ struct to = {0} set all of its members to 0?

---

Back to the answer:

I figured it out: to get it to compile, just delete the zero:

# does NOT work
myStruct _m1 = {0}; 

# works!
myStruct _m1 = {};

It now compiles. However, I ran a bunch of tests to check some things in my struct_initialization.cpp file in my eRCaGuy_hello_world repo, and that does NOT initialize all elements of the struct to zero! Rather, it initializes the struct to its default values. To run my tests and see for yourself, clone my repo above and run eRCaGuy_hello_world/cpp/run_struct_initialization.sh.

Assuming you have this struct:

typedef struct
{
    int num1 = 100;
    int num2 = -100;
    int num3;
    int num4 = 150;
} data_t;

Note: the typedef above is a carry-over from when I was testing this stuff in C instead of C++ (although the default struct values are not allowed in C, of course). For C++, this is preferred instead:

struct data_t
{
    int num1 = 100;
    int num2 = -100;
    int num3;
    int num4 = 150;
};

So please ignore it wherever I unnecessarily use typedef to define the structs below.

Anyway, if I declare one of the above data_t structs, and then do this:

data_t d2 = {};
printf("d2.num1 = %i\nd2.num2 = %i\nd2.num3 = %i\nd2.num4 = %i\n\n",
       d2.num1, d2.num2, d2.num3, d2.num4);

...the output will be:

d2.num1 = 100
d2.num2 = -100
d2.num3 = 0
d2.num4 = 150

And I'm not even sure if d2.num3 is zero because it was initialized to zero or because it was left uninitialized, and that memory location happened to contain zero.

As explained here: https://en.cppreference.com/w/cpp/language/zero_initialization, you can also do this:

myStruct _m1{};

In the example above, this code:

data_t d2{};
printf("d2.num1 = %i\nd2.num2 = %i\nd2.num3 = %i\nd2.num4 = %i\n\n",
       d2.num1, d2.num2, d2.num3, d2.num4);

...would produce output identical to what I showed above.

Even in cases where setting the struct to = {0} DOES work, such as this:

// Does NOT do what I expected! Only sets the FIRST value in the struct to zero! 
// The rest seem to use default values.
data_t d3 = {0};
printf("d3.num1 = %i\nd3.num2 = %i\nd3.num3 = %i\nd3.num4 = %i\n\n",
       d3.num1, d3.num2, d3.num3, d3.num4);

...the output is still not what I expected, as it only sets the first value to zero! (I don't understand why):

d3.num1 = 0
d3.num2 = -100
d3.num3 = 0
d3.num4 = 150

On C-style arrays, however (NOT structs), these semantics work fine. Refer to this answer here (How to initialize all members of an array to the same value?). The following lines, therefore, both set all elements of the C-style array to zero when using C++:

uint8_t buffer[100] = {0}; // sets all elements to 0 in C OR C++
uint8_t buffer[100] = {};  // sets all elements to 0 in C++ only (won't compile in C)

So, after much experimentation, it looks like the following several ways are the ONLY ways to zero-initialize a struct, PERIOD. If you know differently, please comment and/or leave your own answer here.

The only ways possible to zero-initialize a struct in C++ are:

1. Be explicit:

    // C-style typedef'ed struct
    typedef struct
    {
        int num1 = 100;
        int num2 = -100;
        int num3;
        int num4 = 150;
    } data_t;
   
    // EXPLICITLY set every value to what you want!
    data_t d1 = {0, 0, 0, 0};
    // OR (using gcc or C++20 only)
    data_t d2 = {.num1 = 0, .num2 = 0, .num3 = 0, .num4 = 0};
   

2. Use memset() to force all bytes to zero:

    data_t d3;
    memset(&d3, 0, sizeof(d3));
   

3. Set all default values to zero in the first place:

    // C-style typedef'ed struct
    typedef struct
    {
        int num1 = 0;
        int num2 = 0;
        int num3 = 0;
        int num4 = 0;
    } data_t;
   
    // Set all values to their defaults, which are zero in
    // this case
    data_t d4 = {};
    // OR
    data_t d5{}; // same thing as above in C++
   
    // Set the FIRST value only to zero, and all the rest
    // to their defaults, which are also zero in this case
    data_t d6 = {0};
   

4. Write a constructor for the C++ struct

    // 1. Using an initializer list
    struct data
    {
        int num1;
        int num2;
        int num3;
        int num4;
   
        data() : 
            num1(0),
            num2(0),
            num3(0),
            num4(0) {}
    };
   
    data d7; // all values are zero
   
    // OR: 2. manually setting the values inside the constructor
    struct data
    {
        int num1;
        int num2;
        int num3;
        int num4;
   
        data()
        {
            num1 = 0;
            num2 = 0;
            num3 = 0;
            num4 = 0;
        }
    };
   
    data d8; // all values are zero
   

5. Use a struct with no default values, and make your object you create from it static

    typedef struct
    {
        int num1;
        int num2;
        int num3;
        int num4;
    } data_t;
   
    // `static` forces a default initialization of zero for each
    // value when no other default values are set
    static data_t d9;
   

6. So, if you have a struct with non-zero default values, and you want to zero all values, you must do it EXPLICITLY! Here are some more ways:

    // 1. Have a `constexpr` copy of the struct that you use to
    // reset other struct objects. Ex:
   
    struct data
    {
        int num1 = 1;
        int num2 = 7;
        int num3 = -10;
        int num4 = 55;
    };
   
    constexpr data DATA_ALL_ZEROS = {0, 0, 0, 0};
   
    // Now initialize d13 to all zeros using the above `constexpr` struct 
    // object
    data d13 = DATA_ALL_ZEROS; 
   
    // OR 2. Use a `zero()` member function to zero the values:
   
    struct data
    {
        int num1 = 1;
        int num2 = 7;
        int num3 = -10;
        int num4 = 55;
   
        zero()
        {
            num1 = 0;
            num2 = 0;
            num3 = 0;
            num4 = 0;
        }
    };
   
    data d14;
    d14.zero();
   

The big take-away here is that NONE of these: data_t d{}, data_t d = {}, and data_t d = {0}, actually set all members of a struct to zero!

1. data_t d{} sets all values to their defaults defined in the struct.
2. data_t d = {} also sets all values to their defaults.
3. And data_t d = {0} sets only the FIRST value to zero, and all other values to their defaults.

SO, BE EXPLICIT!

Note that the above key take-aways I wrote seem to contradict this documentation on cppreference.com, so it has led me to ask this follow-up question listed just below, which has proven VERY helpful to my understanding!

Going further

1. MOST USEFUL: Follow-up question of mine: Why doesn't initializing a C++ struct to = {0} set all of its members to 0?

References:

1. VERY USEFUL:
   
   1. https://en.cppreference.com/w/cpp/language/zero_initialization
   2. https://en.cppreference.com/w/cpp/language/aggregate_initialization
   3. https://en.cppreference.com/w/cpp/language/value_initialization
2. VERY USEFUL: Initializing all members of an array (not struct) to the same value:
   
   1. How to initialize all members of an array to the same value?
   2. [gcc only] How to initialize all members of an array to the same value?
3. https://github.com/ElectricRCAircraftGuy/eRCaGuy_hello_world/blob/master/cpp/struct_initialization.cpp
   
   1. Clone this repo and run the code yourself with cpp/run_struct_initialization.sh

Related:

1. Initializing default values in a struct
2. *****[my own answer, which demonstrate this sort of struct modification/aggregate member reassignment within any function: leds[0] = {10, 20, 30, 40, 50};] Arduino Stack Exchange: Initializing Array of structs

Initialize/reset struct to zero/null

Define a const static instance of the struct with the initial values and then simply assign this value to your variable whenever you want to reset it.

For example:

static const struct x EmptyStruct;

Here I am relying on static initialization to set my initial values, but you could use a struct initializer if you want different initial values.

Then, each time round the loop you can write:

myStructVariable = EmptyStruct;

Why doesn't initializing a C++ struct to `= {0}` set all of its members to 0?

So, why doesn't initializing a C++ struct to = {0} set all of its members to 0?

Because you are only providing one value, while the class has more than one member.

When you have T t{}; or T t = {} what you are doing is called value initialization. In value initialization, if the object/member does not have a default constructor, or a default member initializer, then the compiler falls back to zero initializing the objec/member. So with

data_t d{}

the value of the members in order would be 100, -100, 0 ,150 and that 0 for num3 happens because it has no default and you did not provide a value in the {} so the compiler falls back to zero initializing num3. This is the same with data_t d = {}. With data_t d = {0} you provide the first element, so num1 is 0, but then like the first two, all of the other members are initialized with their default value if they have one, or zero initialized if they don't, giving you 0, -100, 0, 150 for the member values.

This was a change that happened when C++11 was released and allowed for default member initializers.

---

If your data_t was defined like

typedef struct
{
    int num1;
    int num2;
    int num3;
    int num4;
} data_t;

then data_t d{}, data_t d = {}, data_t d = {0} would all leave you with a zero initialized class since there are no default member initializers and the only value you provide in you braced-init-list (the technical name for {...}) is zero so all members become zero.

Does standard C accept `{0}` as an initializer for any struct?

Yes, this is perfectly valid.

All scalar types (integer types, floating point types and pointer types) accept 0 as an initialiser. This is not a fundamental property of scalar types, but it happens to apply to all of them.

All aggregate types (arrays and structures) and union types will have at least one member or element. Either the first member or element is a scalar type, or it is another aggregate or union type. If the former, that makes {0} a valid initialiser. If the latter, apply the same logic: that too will have at least one member or element. Either that is a scalar type, or it is another aggregate or union type. Keep going. There is no way to have infinitely nested structures, so you'll always end up at a scalar type.

Non-standard language extensions could invalidate some of these assumptions. For instance, a language extension could define strongly-typed enumeration types which do not accept 0 as an initialiser, or empty structures, or zero-length arrays.

initializing struct with {0}

According to cppreference.com

If the number of initializer clauses is less than the number of members [and bases (since C++17)] or initializer list is completely empty, the remaining members [and bases (since C++17)] are initialized [by their default member initializers, if provided in the class definition, and otherwise (since C++14)] by empty lists, in accordance with the usual list-initialization rules (which performs value-initialization for non-class types and non-aggregate classes with default constructors, and aggregate initialization for aggregates). If a member of a reference type is one of these remaining members, the program is ill-formed.

Foo has no default member initializers (int b{0};), so b will be initialized by list-initialization with an empty list, which means value-initialization for non-class types: b = int() // = 0.

initialize struct to zero except one field

does it promise that all other fields are always zero?

Yes. The members that are not explicitly initialized will be zero initialized as part of the designated initializer. You don't even need that 0 there. This:

sampleStruct globalStruct = {.b = true};

should suffice.

If I zero initialize a struct at the start of a loop using {0}, will it be zeroed out every iteration?

Will every compiler actually zero out the struct at the start of every loop?

Any compiler that conforms to the C Standard will do this. From this Draft C11 Standard (bold emphasis mine):

6.8 Statements and blocks

…

3    A block allows a set of declarations and statements to be grouped into
one syntactic unit. The initializers of objects that have automatic
storage duration, and the variable length array declarators of
ordinary identifiers with block scope, are evaluated and the values
are stored in the objects (including storing an indeterminate value in
objects without an initializer) each time the declaration is reached
in the order of execution, as if it were a statement, and within each
declaration in the order that declarators appear.

In the case of a for or while loop, a declaration/initializer inside the loop's scope block is reached repeatedly on each and every iteration of the loop.

What does = { 0 }; mean in a structure initialization in C?

Because of the initialization rules of C, it's kind of a universal initializer:
it will initialize numbers or pointers as well as aggregates (=structs/unions) or arrays by setting everything (every member recursively) to zero.

For block scope objects on platforms where the null pointer is "all bits zero" (most platforms), it's equivalent to memset(&object,0,sizeof(object)); and compilers will often generate such a memset call for {0} initializations, particularly when such initializations are applied to a larger object.

---

Related Topics