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Lets say I have
void swap(int &x, int &y){
//do swap here
}
Why is it legal to do: x = y inside of the function so that it assigns the VALUE of y to x?
Why isn't some sort of dereferencing needed?
There's no dereferencing needed since references are not pointers. They are aliases to the referenced object.
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I have the following code
vector::vector(const vector& arg)
:sz{arg.sz}, elem{new double[arg.sz]}
{
copy(arg, arg+sz, elem);
}
How is arg converted into an iterator here? whats the mechanism called that defines the conversion internally of vector outputting the needed iterator?
Thanks
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vector<char> vec;
vec.push_back(0x1);
char* a = "qwe";
I want to push a to the end of the vector.
You can use the .insert() member function with pointers standing in as iterators.
vec.insert(vec.end(), a, a+strlen(a));
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I am having problems in understanding this statement. I don't know why this is not usual like others.
Assignment means giving a new value to an already existing object. Even though const char INITIAL='G'; has an = sign, it is not an assignment, because it is creating a new object, not modifying an existing one. char INITIAL; INITIAL='G'; would be an assignment, because INITIAL already exists when the new value is, well, assigned.
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I have a matrix in one class. That matrix is allocated dinamically, I want to encapsulate it.
Here is my matrix declaration in the Header file:
float** matrix;
And here is the declaration of get method:
float *getMatrix();
Is everything correct so far?
I don't know how to work with pointers in that case. How would the get function look like?
if you want just to return matrix you should use
float** getMatrix();
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All my attempts so far have failed.Basically when I return a copy of the internal char array of the string, that copy has to be released, but I don't know how to release it.Wrapping it in a smart pointer doesn't work out, since it's destructor gets called immediately after I return it.Must I implement something like a garbage collector just for the immutable string?
const std::string will be fine.