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Why LOG and 0 are used in this statement?
LOG = 35;
vector<int> cnt(LOG, 0); //here cnt is a vector name
The statement is constructing a std::vector, so you should look at help documentation for the vector constructor.
In this case, LOG and 0 are being used specifically in the two-parameter override of the constructor.
LOG is used to specify the initial size of the vector;
0 is used to specify the initial value of the elements.
In other words, the expression declares a vector named cnt being of size 35 with all elements initialized to 0.
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The uninitialized means that the array should firstly be created, but without any value initialized. It will gots values if and only if we do operations on the elements.
For example, given an array A[100]. I want to create another array B[100] with each element in it larger than A by 1, i.e. B[i] = A[i] + 1. But I don't want to initialize all the elements at the beginning. How can I implement such a structure/object?
To be more specific, the operation I want to avoid is the initialization of the elements in B because I need to change the values later. Instead, I want to assign the values directly by some functions/operations.
I think a struct is needed here but I am not familiar with the concepts of the details about how to implement the value assignment once each element got some operations on it.
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Is there any difference in output behaviour between using size(x) instead of x.size(), where x is a string variable? Or it is just another alias?
The behavior of std::size is:
Returns the size of the given container c or array array.
1-2) Returns c.size(), converted to the return type if necessary.
...
So calling std::size(x) and x.size() has the same effect, where x is a std::string.
(This function can also accept an array, but that's not relevant when it comes to a std::string).
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I tried this:
std::map<int,int> m;
and it works -- m becomes an empty map. But this approach may not work if the compiler choose to not initialize m to an empty map by default. Better solution?
Any better solution?
Taking your question literally, no. There is no better solution.
This will create a default constructed, and therefore empty std::map<int,int>.
std::map<int,int> m;
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Let's say I have an array like:
{1,2,3}
So, for example, some function like:
arr.push(4);
Will make this array:
{4,1,2,3}
How do I do that?
It is not possible to push an element to an array. The size of an array remains the same through the lifetime of the array.
What can be done instead is to create a new, larger array and copy the elements from the old array. Such dynamic growable "array" data structure is provided for you in the standard library: std::vector.
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When I print the default value of array on heap memory I'm getting random big numbers in code block. I know that the default value of array is 0 but I am getting random number .
The doc of the std::array constructor says it:
initializes the array following the rules of aggregate initialization (note that default initialization may result in indeterminate values for non-class T)