Why Does My C Code Show an Empty Address for a `char` Member and an 8-Byte Address Difference Between `int` and `string` Members?

Address of Character Data Display Discrepancy

In the provided C code, the output of addresses for class members exhibits an anomaly: the address of the char data member "b" is displayed as an empty space.

The reason for this behavior lies in how the address-of operator and the insertion operator "<<" interpret the char type. When the address of "b" is obtained, it results in a char * pointer. The insertion operator, however, interprets this as a C-style string and attempts to print its contents rather than its address.

To rectify this, the "b" address can be cast to a void pointer using either a C-style cast like "(void ) &b" or, preferably, a safer alternative like "static_cast(&b)". This ensures that the address of "b" is printed correctly.

Mystery of the Address Difference (String and Char)

Another observation is that when the class members int, char, and string are declared as public, the address of "char" is displayed as an empty string, while the difference between the address of "string" and that of "int" is always 8 (not 9).

This difference arises from the way objects are stored in memory. The string data type is implemented as a contiguous block of characters, including a terminating null character. This null character effectively adds an extra byte to the string's size.

Therefore, the difference between the address of "string" and "int" is 8 bytes, representing the size of the string data type plus one for the null-terminator. In contrast, the char data type does not have a null-terminator, resulting in an empty display when its address is printed.

The above is the detailed content of Why Does My C Code Show an Empty Address for a `char` Member and an 8-Byte Address Difference Between `int` and `string` Members?. For more information, please follow other related articles on the PHP Chinese website!