与32位和64位平台的INT64_T匹配的整数字面形式？
考虑下面的一件代码。是否有一个整数字面形式可以在32位和64位平台上编译？
#include <iostream>
#include <cstdint>
void f(double)
{
std::cout << "double\n";
}
void f(int64_t)
{
std::cout << "int64_t\n";
}
int main()
{
f(0L); // works on 64-bit fails on 32-bit system
f(0LL); // fails on 64-bit but works on 32-bit system
f(int64_t(0)); // works on both, but is ugly...
f(static_cast<int64_t>(0)); // ... even uglier
return 0;
}
在平台上int64_t是long，0ll是另一种类型，而Orderload分辨率不希望它与double。
当int64_t是长长（包括在Windows X64上）时，我们与0L有同样的问题。
（0ll是int64_t在32位和X86-64 Windows ABIS（LLP64）中均使用X86-64 System V Windows ABI（LLP64），这是LP64 ABI。当然，非X86系统可移植的东西会很好。）
Consider a piece of code below. Is there an integer literal that would compile on both 32-bit and 64-bit platforms?
#include <iostream>
#include <cstdint>
void f(double)
{
std::cout << "double\n";
}
void f(int64_t)
{
std::cout << "int64_t\n";
}
int main()
{
f(0L); // works on 64-bit fails on 32-bit system
f(0LL); // fails on 64-bit but works on 32-bit system
f(int64_t(0)); // works on both, but is ugly...
f(static_cast<int64_t>(0)); // ... even uglier
return 0;
}
On platforms where int64_t is long, 0LL is a different type and overload resolution doesn't prefer it vs. double.
When int64_t is long long (including on Windows x64), we have the same problem with 0L.
(0LL is int64_t in both the 32-bit and x86-64 Windows ABIs (LLP64), but other OSes use x86-64 System V which is an LP64 ABI. And of course something portable to non-x86 systems would be nice.)
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您可以进行自定义用户定义的文字 for
int64_t喜欢然后您的函数调用将变成
这种情况，如果您尝试使用文字
-92233720368547775808，则会为您带来错误的价值。 /45469214/Why-do-the-the-the-the-the-the-int-in-in-rorr-abour-about-abour-amboud-ambiul-function-ove/45469321＃45469321“>在这里You can make a custom user defined literal for
int64_tlikeand then your function call would become
This will give you an incorrect value if you try to use the literal
-9223372036854775808for the reason stated here