foo(char*)：使用硬编码的 {...}

Question

我有一个函数

foo

需要一个

char*

。我想用硬编码来调用该函数

{0x00, 0xff}

而不先将其分配给变量。

// out of my control, uses char* as a byte array:
void foo(const char*){}

int main(void)
{
  // First assigning to unsigned char (unsigned necessary for 0xff)
  // then casting to char* and calling works:
  unsigned char x[2] = {0x00, 0xff};
  foo((char *)x);

  // foo({0x00, 0xff}); // err, cannot interpret the braces

  // Hint: const char
  foo((const char[]){0x00, 0x01}); // works!

  // 0xff out of char range. Must be unsigned char.
  foo((const char[]){0x00, 0xff});  // err

  return 0;
}

a.cpp:10:6: error: cannot convert ‘<brace-enclosed initializer list>’ to ‘char*’``` 有什么提示吗？

Answer 1

我相信这个

foo(reinterpret_cast<const char*>((const unsigned char[]){0x00, 0xff})); 就可以了。