注意:這與Determine number of bits in integral type at compile time非常相似,但是這是一個非常簡單的版本,在一個.cpp
像msg(long)``和候選`msg(int32_t)`和`msg(int64_t)``
編輯:增加了一個解決方案 - 雖然正確解釋給出(並接受),我發現了一種一般解決問題。
問題問題是與功能類似於
msg(int32_t);
msg(int64_t);
像
long long myLong = 6;
msg(myLong); // Won't compile on gcc (4.6.3), call is ambiguous
電話本上編譯MSVC。任何人都可以提供一個解釋爲什麼在gcc上失敗(我假設它可能與gcc通常嚴格符合標準的事實有關)以及如何正確實現相同效果的示例?
#include <iostream>
#include <stdint.h>
#include <boost/integer.hpp>
using namespace std;
void msg(int v) { cout << "int: " << sizeof(int) << ' ' << v << '\n'; }
void msg(long v) { cout << "long: " << sizeof(long) << ' ' << v << '\n'; }
void msg(long long v) { cout << "long long: " << sizeof(long long) << ' ' << v << '\n'; }
void msg2(int32_t v) { cout << "int32_t: " << sizeof(int32_t) << ' ' << v << '\n'; }
void msg2(int64_t v) { cout << "int64_t: " << sizeof(int64_t) << ' ' << v << '\n'; }
void msg2(uint32_t v) { cout << "uint32_t: " << sizeof(uint32_t) << ' ' << v << '\n'; }
void msg2(uint64_t v) { cout << "uint64_t: " << sizeof(uint64_t) << ' ' << v << '\n'; }
int main()
{
int myInt = -5;
long myLong = -6L;
long long myLongLong = -7LL;
unsigned int myUInt = 5;
unsigned int myULong = 6L;
unsigned long long myULongLong = 7LL;
msg(myInt);
msg(myLong);
msg(myLongLong);
msg2(myInt);
msg2(myLong); // fails on gcc 4.6.3 (32 bit)
msg2(myLongLong);
msg2(myUInt);
msg2(myULong); // fails on gcc 4.6.3 (32 bit)
msg2(myULongLong);
return 0;
}
// Output from MSVC (and gcc if you omit lines that would be commented out)
int: 4 5
long: 4 6
long long: 8 7
int32_t: 4 -5
int32_t: 4 -6 // omitted on gcc
int64_t: 8 -7
uint32_t: 4 5
uint32_t: 4 6 // omitted on gcc
uint64_t: 8 7
解
的解決方案是提供一種成功映射int
,long
和long long
到適當int32_t
或int64_t
的功能。這可以在運行時通過if (sizeof(int)==sizeof(int32_t))
類型語句輕鬆完成,但編譯時解決方案更可取。編譯時解決方案可通過使用boost::enable_if
獲得。
下面的工作在MSVC10和gcc 4.6.3。 解決方案可以通過禁用非整數類型來進一步增強,但這超出了這個問題的範圍。
#include <iostream>
#include <stdint.h>
#include <boost/integer.hpp>
#include <boost/utility/enable_if.hpp>
#include <boost/type_traits/is_signed.hpp>
#include <boost/type_traits/is_unsigned.hpp>
using namespace std;
template <class InputT>
typename boost::enable_if_c<sizeof(InputT)==sizeof(int32_t) && boost::is_signed<InputT>::value,
int32_t>::type ConvertIntegral(InputT z) { return static_cast<int32_t>(z); }
template <class InputT>
typename boost::enable_if_c<sizeof(InputT)==sizeof(int64_t) && boost::is_signed<InputT>::value,
int64_t>::type ConvertIntegral(InputT z) { return static_cast<int64_t>(z); }
template <class InputT>
typename boost::enable_if_c<sizeof(InputT)==sizeof(uint32_t) && boost::is_unsigned<InputT>::value,
uint32_t>::type ConvertIntegral(InputT z) { return static_cast<uint32_t>(z); }
template <class InputT>
typename boost::enable_if_c<sizeof(InputT)==sizeof(uint64_t) && boost::is_unsigned<InputT>::value,
uint64_t>::type ConvertIntegral(InputT z) { return static_cast<uint64_t>(z); }
void msg(int v) { cout << "int: " << sizeof(int) << ' ' << v << '\n'; }
void msg(long v) { cout << "long: " << sizeof(long) << ' ' << v << '\n'; }
void msg(long long v) { cout << "long long: " << sizeof(long long) << ' ' << v << '\n'; }
void msg2(int32_t v) { cout << "int32_t: " << sizeof(int32_t) << ' ' << v << '\n'; }
void msg2(int64_t v) { cout << "int64_t: " << sizeof(int64_t) << ' ' << v << '\n'; }
void msg2(uint32_t v) { cout << "uint32_t: " << sizeof(uint32_t) << ' ' << v << '\n'; }
void msg2(uint64_t v) { cout << "uint64_t: " << sizeof(uint64_t) << ' ' << v << '\n'; }
int main()
{
int myInt = -5;
long myLong = -6L;
long long myLongLong = -7LL;
unsigned int myUInt = 5;
unsigned int myULong = 6L;
unsigned long long myULongLong = 7LL;
msg(myInt);
msg(myLong);
msg(myLongLong);
msg2(ConvertIntegral(myInt));
msg2(ConvertIntegral(myLong));
msg2(ConvertIntegral(myLongLong));
msg2(ConvertIntegral(myUInt));
msg2(ConvertIntegral(myULong));
msg2(ConvertIntegral(myULongLong));
return 0;
}
MSVC有'typedef _Longlong int64_t'和gcc有'typedef long long int int64_t',所以我認爲這兩種情況下的類型是一樣的。無論如何,這是使用_long_的調用是問題... – Zero