How to Print Addresses with Char Pointers Using `cout`?

Cout Interpretation of Char Pointers

Unlike printf(), which offers specific conversion specifiers (%u or %s) to determine whether to print an address or the entire string pointed to by a char pointer, cout requires an explicit approach for this distinction.

Example Problem:

Consider the following code snippet:

<code class="cpp">int main() {
  char ch = 'a';
  char *cptr = &amp;ch;
  cout &lt;&lt; cptr &lt;&lt; endl;
  return 0;
}

In this example, with the default GNU compiler, cout interprets the char pointer as a C-style string and attempts to print the character pointed to by cptr. However, if the intention is to print the address of ch instead, a different approach is necessary.

Solution:

To print the address of ch using cptr with cout, explicit type casting is required. This is achieved by utilizing the static_cast<> operator, as demonstrated below:

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

By explicitly casting cptr to void *, the overload resolution selects the appropriate ostream& operator that takes a void pointer as an argument. This correctly prints the address of ch.

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