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Jonathan Mee
I know that a `const char*` is a pointer to a constant character, and that a `const char[]` is an array of constant characters.

That said what should make me prefer declaring as one type over the other? Are there any good heuristics for directing that decision?
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Jonathan Mee
The simple answer is that when declaring a variable you should prefer a `const char[]`.


To understand this let's take 2 character strings: `const char* ptr = "lorem"` and `const char arr[] = "ipsum"` And lets pass them into this function:

    template <typename T>
    void Output(const void* pv, const char pc[], T ppc) {
        std::cout << "Value: " << pv << " String: " << pc << " Address: " << ppc << std::endl;
`Output(ptr, ptr, &ptr)` will output something like:

> Value: 00E33130 String: lorem   Address: 00FBF8D8

While `Output(arr, arr, &arr)` will output something like:

> Value: 00FBF8DC   String: ipsum   Address: 00FBF8DC

The notable difference here is that `ptr` requires the allocation of another address. There are 3 extra expenses to this.

1. The 1^st^ is the obvious memory cost of a pointer, which one would expect that to be as simple as 8-bytes on a 64-bit architecture, [but because of relocations that can skyrocket to 32-bytes of memory](
1. [The 2^nd^ is the hidden cost of indirection, and potential cache misses](
1. The final expense is in the requirement of separately finding the end of the string pointed to by `ptr`, whereas `arr` can use the [Range Access Functions](

If we were to change `ptr` to a _constant_ pointer (`const char* const ptr`) the result would be the compiler optimizing away **1**, however both **2** and **3** would persist _and_ `ptr` could not be used in functions taking a `const char*`. Ultimately, when declaring variables `const char[]` is the clear choice over both `const char*` and `const char* const`.

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