Why Doesn't the Address of a Char Variable Display Correctly in C ?

Why is Address of Char Data Not Displayed?

Consider this code:

class Address {
public:
    int i;
    char b;
    string c;
    
    void showMap();
};

void Address::showMap() {
    cout <p>When running this code, you may notice that the address of the char variable 'b' is not displayed in the output. Instead, you see a blank area. Why is this happening?</p><p>The reason lies in how the address-of operator (&) works with different data types. When applied to a char variable, &b, you obtain a char pointer, i.e., char *.</p><p>By default, the output operator </p><p>To resolve this issue, you can use the following approach:</p><pre class="brush:php;toolbar:false">cout <p>Here, we cast the address of b to void * before using it with the output operator. This ensures that the operator treats it as a pointer to a memory location, not a character sequence.</p><p><strong>Why is the Difference Between Addresses 8?</strong></p><p>If you make the member variables 'int', 'char', and 'string' public, you may observe another interesting behavior. The difference between the addresses of 'int' and 'string' is consistently 8. This is because:</p>

• The default alignment for int on most architectures is 4 bytes.
• The default alignment for char is 1 byte.
• The default alignment for string is 4 bytes.

When the compiler allocates memory for the class members, it does so in a way that aligns them to their respective boundaries. In this case, the char variable 'b' takes up a single byte, leaving 3 bytes of unused space. This space is then followed by the string variable 'c', which aligns to the nearest 4-byte boundary.

As a result, the difference between the addresses of '&c' and '&b' will be 1 (for 'b') 3 (for unused space) 4 (for the 4-byte alignment of 'c') = 8.

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