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I am a little confused here. In the statement const char *str="Hello";, str is a pointer to a char variable pointing to the first character 'H' of the string "Hello", so str should contain the address of the 'H' character. And yet, if I use cout<<str, it prints the entire string "Hello" and not the address.

And also, If I use cout<<*str to print the value stored in the address pointed to by str, it prints the char 'H' correctly.

Can someone please explain how and why this happens? This may be very basic, but an explanation would help me understand these concepts more clearly.

Remy Lebeau
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Sailesh Khandelwal
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  • `std::cout << static_cast(str);` should do what you want. There are several overloaded `operator<<` for `std::basic_ostream` that take variants of `char *`, which treat the variable like a c-style string. This is done because this is what people expect it to do. https://en.cppreference.com/w/cpp/io/basic_ostream/operator_ltlt2 – JohnFilleau Nov 28 '20 at 18:47
  • There is a specific overload `<<` for C-string, which is better candidate than overload for `void*`. – Jarod42 Nov 28 '20 at 18:53

1 Answers1

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str is a pointer to a char variable

There is no char variable in the example. str does point to a char object.

Can someone please explain how and why this happens?

Because the binary operator << whose left hand operand is an output sream and right hand operand is a pointer to char is specified to print the entire string.

If the pointed char isn't within an array that contains the null terminator character starting from that character, then the behaviour of the program would be undefined. String literals are null terminated, so the example is correct.

eerorika
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