BUG: Fast Compiler Generates Bad Code for Decrement OperatorLast reviewed: July 22, 1997Article ID: Q116168 |
1.00 1.50
WINDOWS
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The information in this article applies to:
SYMPTOMSThe fast compiler may generate bad code for statements that use the decrement operator and may assign the decremented value back to the variable being decremented.
RESOLUTIONThis problem can be worked around either by compiling using the optimizing compiler or by restructuring the code. Use of the optimizing compiler can be forced by using the /f- compiler switch. The sample code in the "MORE INFORMATION" section below can be used both to illustrate the problem and to show how the code can be restructured to work around the problem.
STATUSMicrosoft has confirmed this to be a bug in the C/C++ compiler for MS-DOS, versions 8.0 and 8.0c. We are researching this problem and will post new information here in the Microsoft Knowledge Base as it becomes available. This is not a problem with the 32-bit C/C++ compiler, version 8.0.
MORE INFORMATIONThe following sample code can be used to demonstrate the problem.
Sample Code
/* Compile options needed: /f */ #include <iostream.h> enum color { red, green, blue }; void main() { color ColorEntered; int choice, first = 3, last = 4; do { cout << "\nEnter a 0 or 1. Enter 2 to end: "; cin >> choice; ColorEntered = (color)choice; switch(ColorEntered) { case red: cout << "RED" << endl; break; case green: cout << "GREEN" << endl; // The fast compiler generates bad code for this // statment: first = ( (first > 0) ? (first--) : (last) ); // To work around the problem restructure the code // as follows: /* if( first > 0 ) first--; else first = last; */ cout << "value of first: " << first << endl; break; case blue: cout << "BLUE" << endl; cout << "Ending" << endl; break; default: cout << "Value out of range, ending" << endl; ColorEntered = blue; break; } } while(ColorEntered != blue); } |
Additional reference words: 1.00 1.50 8.00 8.00c
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