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What does int nums[5]; do? When I pass nums to std::cout, it prints a memory address I think, but I don't understand what the code itself is actually doing when it runs.

Remy Lebeau
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Anvita Hodekar
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    You'll want to learn what an *array* is. – Drew Dormann Nov 19 '15 at 02:49
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    Possible duplicate of [What is array decaying?](http://stackoverflow.com/questions/1461432/what-is-array-decaying) –  Nov 19 '15 at 02:51
  • @NickyC: While that answer addresses the question of why `cout` prints a memory address for an array, it does not address the question actually asked here, so I do not think it is a duplicate, just additional useful info. – Remy Lebeau Nov 19 '15 at 02:55

2 Answers2

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int nums[5]; allocates memory for a static array of 5 int values.

When you then do std::cout << nums;, it is actually calling std::cout.operator<<(nums); While std::cout has many << operators defined for many different type types, it does not have an << operator that accepts an int[] array as input. What it does have is an << operator that accepts a const void* memory pointer as input. It prints the value of the memory address that the pointer is pointing at.

A static array can "decay" into a pointer, in this case to an int*.

What is array decaying?

Any type of pointer can be assigned to a void*. And any non-const variable can be assigned to a const variable of compatible type. That is why the compiler does not complain when you call std::cout << nums;. It is essentially acting similar to std::cout.operator<<((void*)(int*)nums); behind the scenes.

Community
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Remy Lebeau
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'nums' is an array that holds 5 int type datas.For Example: int nums[5] = {1,2,3,4,5};

if you want to cout nums,you should write your code like this:

for(int index = 0; index < 5; index ++){
    std::cout<<nums[index]<<std::endl;
}

but ,if you want cout it's memory address,you should write your code like this:

for(int index = 0; index < 5; index ++){
    std::cout<<nums<<std::endl;
    nums ++;
}
yditxu
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