can somebody explain me why it's possible to do:
String s = "foo";
how is this possible without operator overloading (in that case the "=")
I'm from a C++ background so that explains...
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In this case there is no overloading. The java piece that differs from C++ is the definition of "" - The java compiler converts anything in "" into a java.lang.string and so is a simple assignment in your example. In C++ the compiler converts "" into a char const * and so needs to have a conversion from char const* to std::string.
mmmmmm
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Samuel what about it - the question is in effect why is there no implicit or explicit conversion as there woud be in C++ for String s = "foo"; - how would + come in here? – mmmmmm Feb 24 '10 at 17:24
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The compiler converts the series of concatentations into a StringBuffer instantiation followed by append calls with the String objects as mentioned in this answer. http://java.sun.com/developer/JDCTechTips/2002/tt0305.html – basszero Feb 24 '10 at 17:26
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Ah I see - string concatination is also detailed in the Java language docs http://java.sun.com/javase/6/docs/api/java/lang/String.html – mmmmmm Feb 24 '10 at 17:34
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`String` does overload `+`—the one and only case of operator overloading in Java. – Taymon Jun 01 '12 at 23:09
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It is an assignment operator in java which is used to assign the value to the declared type, where no operator overloading is required. Even in c++
GuruKulki
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"Strings are constant; their values cannot be changed after they are created. String buffers support mutable strings. Because String objects are immutable they can be shared. For example:
String str = "abc";
"is equivalent to:
char data[] = {'a', 'b', 'c'};
String str = new String(data);
David
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