When I want to print the address of any char variable then, output will be a single char or blank.
i.e,
char x = 'a';
cout<<&x;
output:
f
Please! tell me why is is happening...?
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DevSolar
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Sachin Kadian
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This is because there's an overload for this specific pointer type. – Hatted Rooster Jan 13 '22 at 12:26
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Please suggest me the solution of this. – Sachin Kadian Jan 13 '22 at 12:27
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Try this: `cout<
(&x);` – Eljay Jan 13 '22 at 12:28 -
Since address doesn't depend on the underlying type, you can cast it to a pointer to any type and print it. I think this should work `(void*)(&x)` – Alexey S. Larionov Jan 13 '22 at 12:28
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what is the function of static_cast ?And void * . – Sachin Kadian Jan 13 '22 at 12:31
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[`static_cast`](https://en.cppreference.com/w/cpp/language/static_cast). `void*` is a pointer that points to any kind object or nullptr (or dangling pointer or wild pointer or uninitialized pointer or garbage pointer), where `void` is used kind of as a wildcard/placeholder in lieu of the type. – Eljay Jan 13 '22 at 12:45
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@SachinKadian please [read a good C++ book](https://stackoverflow.com/q/388242/995714) which explains everything – phuclv Jan 13 '22 at 12:57