C用户定义的文字运算符是否可以传递空指针？
这实际上发生在g的实验版本(gcc版本4.7.0 20111114(实验性)[主干版本181364](Debian 20111114-1))但我不确定这是否是一个错误(90％肯定)或一些奇怪的预期行为.
示例程序：
#include <iostream>
#include <stdexcept>
#include <string>
std::string operator "" _example (const char * text) {
using std::cerr;
using std::endl;
cerr << "text (pointer) = " << static_cast<const void *>(text) << endl;
if (!text) throw std::runtime_error("Is a null pointer really expected?");
cerr << "text (string) = \"" << text << '"' << endl;
return text;
}
int main() {
2_example;
1_example;
0_example;
}
输出(可能是gcc中的一个bug ……但也许不是？！,因此问题)：
text (pointer) = 0x8048d49 text (string) = "2" text (pointer) = 0x8048d4b text (string) = "1" text (pointer) = 0 terminate called after throwing an instance of 'std::runtime_error' what(): Is a null pointer really expected? Aborted
它不仅仅是“0_example”;每当字面值为零时都是如此.例如,即使文字是“0x0000_example”,它仍然会发生.
