c++ right way to initialize a vector of a struct - c++

Question

i searched a lot here, but there is no right explanation for me, for an advanced newbie in c++. I worked before with vector of structs and now I get segmentation faults...
Thats why I want to know how such objects actually works and if it is the right the way I am doing!
I have a struct like
struct numberOfSpecies {
int predator;
int prey1;
int prey2;
};
and a vector of it:
std::vector<numberOfSpecies> size;
Before I resize it and fill it with values.
size.resize(100);
what is actually this doing? Is this right for a struct?
It looks like it is initialized with zeros...
Now I am doing this like:
size[t].predator=0;
size[t].prey1=0;
size[t].prey2=0;
for(int k = 0; k < N; ++k){
size[t].predator++;
size[t].prey1++;
size[t].prey2++;
}
Is this right? Where are possible issues? How to do it better?

Answer 1

The easiest and 'correct' solution here is probably to just use the resize() function that belongs to the vector object with aggregate initialization (if you have access to c++11 and on), something like
size.resize(100,{0,0,0}); //aggregate initialization
for(int k = 0; k < N; ++k)
{
size[t].predator++;
size[t].prey1++;
size[t].prey2++;
}
All members of each numberOfSpecies object will be initialized to 0.

Answer 2

This:
size[t].predator=0;
size[t].prey1=0;
size[t].prey2=0;
will write zeros to the tth element of size - that may or may not be useful:
This:
for(int k = 0; k < N; ++k){
size[t].predator++;
size[t].prey1++;
size[t].prey2++;
}
will increment the tth element of size N times. This seems incredibly unlikely to be useful. I think what you want is:
size[0].predator=0; // Technically not needed because .resize()
size[0].prey1=0; // will have initialized it to zero anyway
size[0].prey2=0; // *BUT* explicit is always better than implicit.
// Initialize each element of size to be one greater than previous.
for(int k = 1; k < N; ++k){
size[k].predator = size[k-1].predator + 1;
size[k].prey1 = size[k-1].prey1 + 1;
size[k].prey2 = size[k-1].prey2 + 1;;
}

Answer 3

Use the value parameter for static parameters.
#include <vector>
struct foo{
int g;
int h;
int l;
};
int main()
{
std::vector<foo> manyFoo(10, {0});
manyFoo.resize(60, {0});
}
If you want to grow your vector as you also put arbitrary values into the struct you could ->
#include <iostream>
#include <vector>
struct foo{
foo(int aG,int aH, int aL):g(aG),h(aH),l(aL) {};
int g;
int h;
int l;
};
int main() {
std::vector<foo> lVec;
for (int i=0;i<10;i++) {
lVec.emplace_back(foo(i,i*2,i*4));
}
int lPos=0;
for (auto &rFoo: lVec) {
std::cout << "Item pos" << lPos++ << " g:" << rFoo.g << " h:" << rFoo.h << " l:" << rFoo.l << std::endl;
}
return EXIT_SUCCESS;
}
If you know the size of the vector and you want to populate it you could ->
#include <iostream>
#include <vector>
struct foo{
foo(int aG,int aH, int aL):g(aG),h(aH),l(aL) {};
int g;
int h;
int l;
};
int main() {
std::vector<foo> lVec(10,{0,0,0});
int lPos = 0;
for (auto &rFoo: lVec) {
rFoo = foo(lPos,lPos*2,lPos*4);
lPos++;
}
lPos=0;
for (auto &rFoo: lVec) {
std::cout << "Item pos" << lPos++ << " g:" << rFoo.g << " h:" << rFoo.h << " l:" << rFoo.l << std::endl;
}
return EXIT_SUCCESS;
}

Answer 4

You could add an default constructor to your structure. The new code will look something like this:
struct numberOfSpecies {
numberOfSpecies (): predator(0), prey1(0), prey2(0) { } // default constructor
int predator;
int prey1;
int prey2;
};
That way, your structure will be properly initialized inside your vector when resize is applied.