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Cat c6 4 acert engine

Cat c6 4 acert engine


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Cat c6 4 acert engine 4.2 8

3cat c5 4 mack engine 4.2 8

1cat c5 3.9 dfe 11

2cat c5 4 mack engine 4.2 8

2cat c5 3.9 dfe 11

3cat c6 4.2 8

4cat c6 4.2 8

3cat c6 4.2 8

4cat c6 4.2 8

4cat c6 4.2 8

Here is what I'm trying to do with my code:

for(cat = 0, cat<,4, cat++){

for(i=0, i<,5, i++){

cout <,<, cat <,<, ":" <,<, " | " <,<, i <,<, " : " <,<, " " <,<, (i+1) <,<, " | " <,<, my_array[i] <,<, " | " <,<, my_array[i+1] <,<, endl,

}

}

This is the error I'm getting:

error: non-const lvalue reference to type 'std::array<,unsigned short,

1ul>,' cannot bind to a temporary of type 'const unsigned short *'

I tried declaring the array my_array const but I still get this error.

How can I make my array constant and iterate over it without changing my array?

A:

I tried declaring the array my_array const but I still get this error

Because of the problem which @John Fultz and @Krishna identified, you cannot take reference to a pointer and change it without changing its value. You need to copy the array's elements before altering it and return the copy.

How can I make my array constant and iterate over it without changing my array?

Use std::array<,...>,::cbegin(), std::array<,...>,::cend(),

std::array<,...>,::data(), std::array<,...>,::at(index), std::array<,...>,::begin(), std::array<,...>,::end() or std::array<,...>,::value_type{}.

template<,std::size_t SIZE>,

std::array<,unsigned short, SIZE>, cat_array(unsigned short*&, my_array){

unsigned short* cat = my_array,

for(int i=0, i<,5, i++){

cat[i] = 10,

}

return std::array<,unsigned short, SIZE>,{

cat

},

}

If we declare a pointer to a array and try to change that array's elements using pointer arithmetic, the behavior is implementation-defined. It is undefined whether or not the object's original values are kept. The following example illustrates it:

int* p,

int const my_array[] = {0,1,2,3,4},

const int my_const_array[] = {0,1,2,3,4},

p = &,my_array,

*p = 5, // This works.

p = &,my_const_array,

*p = 5, // This doesn't.

A:

The answer to this question depends on how the pointer gets modified.

If the pointer is passed to a function by value, the pointer value is copied to the function argument, which is then modified. So all pointers in the array get modified. This modification affects the pointer, which now points to a different object.

If the pointer is passed by reference, the pointer itself is passed. Thus the pointer variable in the function can access the original memory. It can't modify the contents of that memory however, as it only refers to the memory as the argument value.

A:

Suppose you have the code like this:

int *array[5],

for(int i = 0, i <, 5, ++i)

array[i] = i,

As you know, this code, for instance, when compiled to assembly, the compiler will do some optimization like inline function or loop unrolling. The compiler can change the value of the array in-place.

If you have a const int *array[5],

This does not apply to the function like this:

void foo(const int *a)

{

int *b = a,

b[1] = 9,

}

because in this example, the compiler can't inline the function body and have to make a copy of a. But the compiler may inline the first for-loop of your code.

If you have a const int *array[5], and the pointer is passed by value, then the copy-constructor or move-constructor is always invoked when you are making a function call. The compiler may inline the first for-loop, too.

If you have a const int *array[5], and the pointer is passed by reference, then there will be a copy of the array. You may want to consider passing the address of the array.

As you can see, the compiler will use different rules depending on the function call context, which is why the return type of a function must be fully defined so that the compiler can make the right choice. The C standard says the following in the 6.9.3.2 paragraph 6:

When a function is called (6.2.2), unless the function is a

member of a class with virtual functions, the called function

is guaranteed to have a dedicated stack frame.

So in your particular code, you may invoke foo multiple times, which will make sure that you will have a copy of the array. But this won't happen in the code that you posted in the question, and thus the array will not have any value set.

And in other cases, the caller may use the function for read-only operations, and if the compiler inlines your function, the array may become read-only. For example, if you do


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