C++'s const, constexpr, consteval

Table of contents

The const keywords in C++

C++ is highly regarded for its control over performance. The const keywords are undoubtedly important in achieving optimized applications. These keywords are some of the fundamentals every C++ developer has to know.

In this blog, we are we considering these keywords:

I have yet to come up with a good use for constinit (even cppreference has no examples). That is why we are not discussing it right now. Probably I will have a part 2, or even update this very own blog as well for it.

Immutable vs. Immediate

Before any further explanation, I want to define some terminologies first.

Immutable: after creation, value cannot be changed during and after running.

Immediate: immutable, but value is also known before running.

From the above definition, we can also see that:

An immutable variable can be initialized during either compile time or run time.

An immediate variable can only be initialized during compile time.

It can be seen that immediate variables have stricter conditions than immutable ones. Thus, it allows greater (compile-time) optimizations.

const vs. constexpr variable

Okay, I hope we are settled on the definitions. Now to the actual C++ part.

const

A const variable is immutable.

What this means is that this code is valid:

void valid() {
    int a;
    std::cin >> a;

    const int b = a + 1; // OK

    b = 69; // not OK, b is immutable
}

Let's do a checklist:

The first condition satisfies the definition of an immutable variable.

The second condition doesn't affect its immutability at all. However, it will be important in the next part.

constexpr

A constexpr variable is immediate.

A valid example:

void valid() {
    constexpr int a = 6; // OK, a = 6
    constexpr int b = a + 9; // OK, b = 15

    b = 7; // not OK, b is immediate/immutable
}

The first condition satisfies the definition of an immediate (which is also immutable) variable.

The second condition satisfies the definition of an immediate. The program knows the value a and b before running, and they won't change during and after running.

And an invalid example:

void invalid() {
    int a;
    std::cin >> a;

    constexpr int b = a + 1; // not OK, b is not known before running
}

Thus, it fails to become a constexpr variable.

const number == constexpr?

But, what about this example:

int main() {
    const int b = 10;
}

b now satisfies both conditions of an immediate variable, but it is declared as const. It is actually okay. A good compiler may be able to detect this is effectively the same as a constexpr variable and optimize it.

However, you are still betting on the compiler to do the magic work for you. If you want to be certain, just use constexpr.

What can constexpr variables do?

I'm running the examples in Compiler Explorer for the assembly.

To avoid other sorts of optimizations from the compiler, I'm not using any optimization flags for these examples.

No constexpr

Let's consider an example with no constexpr:

int func1() {
    int a = 17;
    int b = 9;
    return a / b;
}

A simple function. Let's look at the assembly:

func1():
        push    rbp
        mov     rbp, rsp
        mov     DWORD PTR [rbp-4], 17
        mov     DWORD PTR [rbp-8], 9
        mov     eax, DWORD PTR [rbp-4] # load dividend
        cdq
        idiv    DWORD PTR [rbp-8]      # load divisor -> divide
        pop     rbp
        ret

Both the values of a and b are loaded to the memory at [rbp-4] and [rbp-8], respectively.

Note: Due to keeping the stack as is (because no optimization flags are enabled), these two mov instructions are kept throughout all examples. They are not really relevant to our investigation.

For this example:

Full of slow stuff. We can do better.

constexpr dividend

int func2() {
    constexpr int a = 17;
    int b = 9;
    return a / b;
}

func2():
        push    rbp
        mov     rbp, rsp
        mov     DWORD PTR [rbp-4], 17
        mov     DWORD PTR [rbp-8], 9
        mov     eax, 17 # load dividend as 17,
                        # no memory access
        cdq
        idiv    DWORD PTR [rbp-8] # load divisor -> divide
        pop     rbp
        ret

Mostly the same as the previous example. However, there is a difference: eax is now loaded with an immediate value of 17, instead of loading from the memory.

The equivalent C++ code will be:

int func2p() {
    int b = 9;
    return 17 / b;
}

Notice how a is gone. Using constexpr like in func2(), we can have the code with the same performance as func2p() without replacing the variable with its value by hand.

constexpr divisor

int func3() {
    int a = 17;
    constexpr int b = 9;
    return a / b;
}

func3():
        push    rbp
        mov     rbp, rsp
        mov     DWORD PTR [rbp-4], 17
        mov     DWORD PTR [rbp-8], 9
        mov     eax, DWORD PTR [rbp-4] # load dividend
        movsx   rdx, eax
        imul    rdx, rdx, 954437177    # multiply the dividend
                                       # with (1/9 mod 2^32)
        shr     rdx, 32     # these lines of code are
        sar     edx         # normalizing the result
        sar     eax, 31     # of the division
        sub     edx, eax    # to account for inaccuracy
        mov     eax, edx
        pop     rbp
        ret

Notice we are not using idiv anymore. We are using another trick instead.

This seemingly "magic" value of 954437177 is actually (1/9 mod 2^32). Instead of doing 17/9, we are doing 17 * 1/9 with the faster imul instruction. The compiler calculates 1/9 beforehand.

This technique is called "Division by Invariant Multiplication". A bunch of arithmetic instructions in the second half of the function are normalizing the result to compensate for the inaccuracy that may happen.

The equivalent C++ code will be:

int func3p() {
    int a = 17;
    return a / 9;
}

Both constexpr

int func4() {
    constexpr int a = 17;
    constexpr int b = 9;
    return a / b;
}

func4():
        push    rbp
        mov     rbp, rsp
        mov     DWORD PTR [rbp-4], 17
        mov     DWORD PTR [rbp-8], 9
        mov     eax, 1    # return 1
        pop     rbp
        ret

With both sides being constexpr, the compiler just calculates 17/9 == 1, and return 1. That's it.

The equivalent C++ code will be:

int func4p() {
    return 17 / 9; // or: return 1;
}

constexpr function

You have seen what magic those C++ compilers can do with constexpr variables. Let's make a step further and apply constexpr to functions.

However, there is a rather "weird" trait:

A constexpr function can evaluate either during compile time or run time.

The next example can be viewed on Compiler Explorer here.

constexpr int add_five(int n) {
    return n + 5;
}

int main() {
    int n = 5;
    const int a = add_five(n);
    constexpr int b = add_five(10);
}

add_five(int):
        push    rbp
        mov     rbp, rsp
        mov     DWORD PTR [rbp-4], edi
        mov     eax, DWORD PTR [rbp-4]
        add     eax, 5
        pop     rbp
        ret
main:
        push    rbp
        mov     rbp, rsp
        sub     rsp, 16
        mov     DWORD PTR [rbp-4], 5    # n := 5
        mov     eax, DWORD PTR [rbp-4]  # eax := n
        mov     edi, eax                # edi := eax
        call    add_five(int)           # eax := add_five(edi)
        mov     DWORD PTR [rbp-8], eax  # a := eax
        mov     DWORD PTR [rbp-12], 15  # b := 15
        mov     eax, 0
        leave
        ret

add_five is a constexpr function. It is kept in the assembly, because later on, it will be used in run time.

It can be seen that a is an immutable variable as its value will be determined during run time. There is indeed a call to add_five, with n as the argument.

However, b is an immediate variable, as it is constexpr. add_five(10) == 15, and b is assigned with the value 15 without any call to the function.

We can see that for this constexpr function, if it can evaluate during compile time, it will. Otherwise, it will evaluate during run time for other cases.

consteval function

C++20 introduces a new keyword: consteval. It is a keyword used with functions (and not with variables). It aims to provide another tool for compile-time control.

A consteval function can only evaluate during compile time. Otherwise, the program fails to compile.

We have a similar example, but this time using consteval:

consteval int add_five(int n) {
    return n + 5;
}

int main() {
    // the following 2 lines lead to compilation errors:
    // int n = 5;
    // const int a = add_five(n);

    constexpr int b = add_five(10);
}

main:
        push    rbp
        mov     rbp, rsp
        mov     DWORD PTR [rbp-4], 15   # b := 15
        mov     eax, 0
        pop     rbp
        ret

Pretty much the same effect. However, it will fail to compile if it can't evaluate during compile time.

The function add_five is also gone in the assembly as well: it can only be executed in compile time, and it should not exist after compilation is done.

if constexpr

I was watching Carl Cook's talk at CppCon 2017. He showed an example of eliminating the branching away with template specialization.

Scrolling down the comments, there was one mentioning if constexpr. And I thought it was brilliant.

Branching approach

This is the "branching approach" in Cook's talk:

enum class Side { Buy, Sell };

float CalcPrice(Side side, float value, float credit) {
    return side == Side::Buy
         ? value - credit
         : value + credit;
}

It can be pretty fast, but we can go further.

Template specialization approach

Cook's proposal was:

enum class Side { Buy, Sell };

template<Side side>
float CalcPrice(float value, float credit);

template<>
float CalcPrice<Side::Buy>(float value, float credit) {
    return value - credit;
}

template<>
float CalcPrice<Side::Sell>(float value, float credit) {
    return value + credit;
}

There is absolutely no branching. The code is deterministic. It's either on the buy side or the sell side. No other sides.

Let's see what we can do with if constexpr.

if constexpr approach

enum class Side { Buy, Sell };

template<Side side>
float CalcPrice(float value, float credit) {
    if constexpr (side == Side::Buy) {
        return value - credit;
    } else {
        return value + credit;
    }
}

It looks like there is branching with if. However, the branching is done in compile time. The generated assembly will either return value - credit or value + credit. There will be no branch at all. Effectively, we have achieved the same code as the template specialization method, with a centralized, more compact and readable code.

Any other examples?

This StackOverflow answer demonstrates how if constexpr can be different from if. Their example is more on the functionality side, while Carl Cook's example is on the performance side.

if consteval

Do you remember that a constexpr function may execute either during compile time or run time?

if consteval in C++23 will allow us to branch, depending on whether that constexpr function is evaluating during compile time or run time.

constexpr int f() {
    if consteval {
        return 69;
    } else {
        return 420;
    }
}

int main() {
    int a = f(); // a = 420
    constexpr int b = f(); // b = 69
}