Why Does the Behavior of `int64_t`, `long int`, and `long long int` Differ in 32-bit and 64-bit Compilers?

long long int vs. long int vs. int64_t in C

Consider the following C program:

<code class="cpp">#include <iostream>
#include <cstdint>

template <typename T>
bool is_int64() { return false; }

template <>
bool is_int64<int64_t>() { return true; }

int main()
{
    std::cout << "int:\t" << is_int64<int>() << std::endl;
    std::cout << "int64_t:\t" << is_int64<int64_t>() << std::endl;
    std::cout << "long int:\t" << is_int64<long int>() << std::endl;
    std::cout << "long long int:\t" << is_int64<long long int>() << std::endl;

    return 0;
}</code>

In both 32-bit GCC and 32/64-bit MSVC compiles, the output of the program will be:

int:           0
int64_t:       1
long int:      0
long long int: 1

However, in a 64-bit GCC compile, the output will change to:

int:           0
int64_t:       1
long int:      1
long long int: 0

This behavior arises because in 64-bit compiles, int64_t is defined as long int, not long long int. To fix this, we can use platform-specific checks as follows:

<code class="cpp">#if defined(__GNUC__) && (__WORDSIZE == 64)
template <>
bool is_int64<long long int>() { return true; }
#endif</code>

However, this solution is not ideal, as it relies on specific compiler and platform combinations. A more reliable method would be to explicitly specify the type equivalence to the compiler. Unfortunately, C does not provide a way to define such equivalence between basic data types. Instead, we can rely on type traits, such as std::is_same. For example:

<code class="cpp">// C++11
template <typename T>
void same_type(T, T) {}

void foo()
{
    long int x;
    long long int y;
    same_type(x, y); // Will now compile successfully
}</code>

In this example, std::is_same is used to determine whether x and y have the same type, regardless of their underlying representation.

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