c++ - 为什么 char 在数组中占用的空间似乎比它本身占用的空间更多

Question

char testChar = 'a';
char myCharString[] = "asd";
char *pointerToFirstChar = &(myCharString[0]);
char *pointerToSecondChar = &(myCharString[1]);
cout << "A char takes " << sizeof(testChar) << " byte(s)";
cout << "Value was " << pointerToFirstChar << ", address: " << &pointerToFirstChar << endl;
cout << "Value 2 was " << pointerToSecondChar << ", address:" << &pointerToSecondChar << endl;

这个输出:

"A char takes 1 byte"

"... address: 00F3F718"

"... address: 00F3F70C",

我认为地址之间的差异应该是 1 个字节，因为那将是分隔它们的数据的大小。为什么不是这样？

Answer 1

&pointerToFirstChar 和 &pointerToSecondChar，你获取的不是 char 数组元素的地址，而是局部变量的地址pointerToFirstChar 和 pointerToSecondChar。请注意，它们本身就是指针。

你可能想要:

cout << "Value was " << pointerToFirstChar << ", address: " << static_cast<void*>(pointerToFirstChar) << endl;
cout << "Value 2 was " << pointerToSecondChar << ", address:" << static_cast<void*>(pointerToSecondChar) << endl;

请注意，您需要将它们转换为 void* 以打印出地址而不是字符串。