Why does `cout` print the character instead of the address when working with char pointers?

Understanding Char Pointer Printing with cout

When dealing with char pointers, it's crucial to understand how they are handled in different environments. While print() with %u and %s conversion specifiers gives control over printing addresses or strings, the challenge arises when using cout.

Consider the following code snippet:

<code class="cpp">char ch = 'a';
char *cptr = &ch;
cout << cptr << endl;

Here, the desired output is the address of ch stored in cptr. However, by default, cout will tend to treat cptr as a regular string and print the character 'a' instead of its address.

Resolving the Printing Dilemma

To rectify this issue, we need to force cout to interpret cptr as a void pointer rather than a char pointer. This allows us to leverage the ostream& operator that specifically handles void pointers and prints their addresses.

To achieve this, we can utilize a cast as follows:

<code class="cpp">cout << static_cast<void *>(cptr) << endl;</code>

By casting cptr to a void pointer, the overload resolution selects the correct operator, resulting in the desired address printing.

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