CODEWITHSUNDEEP
c++11
Suslik
Suslik

Reputation: 1071

Is there any difference between std::vector::resize and direct initialization to zero

Is there any difference between:

std::vector<int> B;
B.resize(N);

and

std::vector<int> B(N); //or std::vector<int> B(N,0);

or is the second method the short form of the first method?

Upvotes: 0

Views: 39

Answers (1)

Cinder Biscuits
Cinder Biscuits

Reputation: 5251

Is there a measurable difference in performance? No, not at all.

Is there a readability difference? The allocator version is much more concise and easier to grok at a glance.

Is there a difference in the generated instructions? Depends on the compiler and its opts. Let's take a look with gcc:

int main() {
    std::vector<int> B(1);
}

compiles to:

    call    std::allocator<int>::allocator() [complete object constructor]
    lea     rdx, [rbp-17]
    lea     rax, [rbp-48]
    mov     esi, 1
    mov     rdi, rax
    call    std::vector<int, std::allocator<int> >::vector(unsigned long, std::allocator<int> const&)
    lea     rax, [rbp-17]
    mov     rdi, rax

The other case:

int main() {
    std::vector<int> B;
    B.resize(1);
}

Compiles to:

    call    std::vector<int, std::allocator<int> >::vector() [complete object constructor]
    lea     rax, [rbp-48]
    mov     esi, 1
    mov     rdi, rax
    call    std::vector<int, std::allocator<int> >::resize(unsigned long)
    lea     rax, [rbp-48]
    mov     rdi, rax

They are very very similar. The difference is negligible. Only difference here is the allocator version has an additional LEA instruction for loading std::allocator

(LEA does memory addressing calculations, but doesn't actually address memory.) Optimizer fudge.

Bear in mind, compiler optimizations are not standard. As far as the language itself dictates, there would be no difference between these two calls at all.

Upvotes: 2

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