Funny C++

[Almindra]

Who can explain : char *m="az"; cout << m << endl; i=0; cout << m[i++] << m[i++] << endl ; i=0; ((cout.operator<<(m[i++])).operator <<( m[i++])).operator <<(endl) ; Gives (Visual C++) 6.0 : az za za Thanks .

[David Crow]

Using more than one ++ or -- operator on anything but a line by themselves is always risky. Looking at the statement:

cout << m[i++] << m[i++] << endl;

the i++ statement gets executed first, and that value (1) is used with the [] operator.

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"The pointy end goes in the other man." - Antonio Banderas (Zorro, 1998)

[jbarton]

In the second and third cout statements, you are modifying the value of i more than 1 time between sequence points. The result of this is undefined behavior - the compiler can output anything it wants.

[Almindra]

sorry but the value of i used by the [] operator will be 0 ... I think you're confused whith ++i ...

[Tim Smith]

Um, no. The problem here is the undefined behavior of the multiple ++ operations which does get you into some problems. int i = 0; cout << m[i++] << endl; This code would fetch the 0'th value. The ++ and -- operators are not risky when you know how to use them. They have well defined behaviors and undefined behaviors. Tim Smith I'm going to patent thought. I have yet to see any prior art.

[David Crow]

cout << i++ << i++ << endl;

Stepping into the code for ostream& ostream::operator<<(int n) revealed that n had a value of 1 the first time, and a value of 0 the second time. Had the pre-increment operator been used instead, a subscript-out-of-range error would have been realized since the value of n would have been 2 the first time through. Separating the i++ operations into two statements solves all the problems.

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"The pointy end goes in the other man." - Antonio Banderas (Zorro, 1998)