c++ - 使用 new 动态分配字符是否需要编译器将它们初始化为零

Question

例如:

char *p=new char[100];

必须根据 C++ 标准将 p 指向的字符数组初始化为零吗？或者，这种行为是否完全依赖于编译器？

gcc 似乎在每个字符上调用默认构造函数，这当然会将它们初始化为零。 Visual C++ 2010 没有。

Answer 1

不，POD 类型在由 new 创建时未初始化。如果需要，您可以将它们值初始化为零:

char * p = new char[100]();
                        ^^

这是由标准规定的:

C++11, 5.3.4/15: If the new-initializer is omitted, the object is default-initialized (8.5); if no initialization is performed, the object has indeterminate value.

8.5/6: To default-initialize an object of type T means:
— if T is a (possibly cv-qualified) class type (Clause 9), the default constructor for T is called (and the initialization is ill-formed if T has no accessible default constructor);
— if T is an array type, each element is default-initialized;
— otherwise, no initialization is performed.

你的其他观察:

gcc seems to call the default constructor on each character

它不应该这样做 - 特别是因为 char 没有构造函数。如果您将其替换为具有默认构造函数的类型，那么将为每个元素调用。

我得到以下反汇编，没有任何初始化的迹象:

int main() {
    char * p = new char[100];
    return p[0];
}

00000000004005f4 <main>:
  # set up stack frame
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp

  # call `operator new[]` with an argument of 100
  mov    $0x64,%edi
  callq  4004e0 <operator new[](unsigned long)@plt>

  # put the return value into %eax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax

  # return
  leaveq 
  retq