Vector of Vector 如何在内存中对齐? [英] How is a Vector of Vector aligned in memory?

问题描述

我理解为

所以你的

ref.resize(i);

仅影响顶行.你的

ref[i].push_back(23);

只影响第 i 列.

I understand as Set size of vector of vectors at run time describes, one can declare vector of vector as

vector<vector<int> > ref;

then resize the first level by

ref.resize(i);

and push element at the 2nd level:

ref[i].push_back(23);

But how are vector of vector aligned in memory?

For simple vector, it's a container and align its element continuously, like an array; but in the case of vector of vector, I couldn't see the picture.

As the size of each inner vector (the vector in vector of vector) size might change, does the outer vector of vector (the vector in vector of vector) align inner vectors continously? Does the outer vector researve memeory space for each inner vector? what if one vector overshoot?

解决方案

The size of the vector<int> struct that is stored in ref is constant. Common implementations has this as three pointers, or around 12 bytes on 32-bit architectures, or 24 bytes on shiny new 64-bit architectures.

So ref manages roughly ref.capacity() * 12 bytes of continuous storage.

Each element/vector<int> in ref manages its own integers independent of the elements ref manages. In the artistic rendering below ref.size() == ref.capacity() for the sake of simplicity.

So your

ref.resize(i);

only affects the top row. Your

ref[i].push_back(23);

only affects the i-th column.

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