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Floating point representation and output seems to be compiler dependent?
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Answer / rohit sah
Regrettably, yes. The action of ios::setprecision() varies among compilers.
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how can i access a direct (absolute, not the offset) memory address? here is what i tried: wrote a program that ask's for an address from the user, creates a FAR pointer to that adress and shows it. then the user can increment/decrement the value in that address by pressing p(inc+) and m(dec-). NOW, i compiled that program and opened it twice (in 2 different windows) and gave twice the same address to it. now look what happen - if i change the value in one "window" of the program, it DOES NOT change in the other! even if they point to the same address in the memory! here is the code snippet: //------------------------------------------------------ #include <stdio.h> //INCLUDE EVERY KNOWN HEADER FILE #include <conio.h> //FOR ANY CASE... #include <iostream.h> #include <dos.h> #include <process.h> main() { int far *ptr; //FAR POINTER!!! long address; char key=0; //A KEY FROM THE KEYBOARD int temp=0; clrscr(); cout<<"Enter Address:"; cin>>hex>>address; //GETS THE ADDRESS clrscr(); (long)ptr=address; temp=*ptr; //PUTS THE ADDRESS IN THE PTR cout<<"["<<hex<<(unsigned long)ptr<<"]="<<*ptr<<" = "<< (char)(*ptr); //SHOWS: [address]=value=ASCII symbol. while (key!=27) //WHILE YOU DONT PRESS ESC. { while(!kbhit()) //WHILE KEY IS NOT PRESSED { if (temp!=*ptr) { temp=*ptr; clrscr(); cout<<"["<<hex<< (unsigned long)ptr<<"]="<<*ptr<<" = "<<(char)(*ptr); }; //IF THE VALUE HAS CHANGED, CLEAR THE SCREEN AND SHOW //AGAIN if (key=='p') {key=0; (*ptr)++; } //INCREMENT VALUE if (key=='m') {key=0; (*ptr)--; } //DEC. VALUE }; key=getch(); //IF A KEY IS PRESSED, READ IT FROM THE //KEYBOARD }; return 0; //IF ESC WAS THE KEY, EXIT THE PROGRAM } //---------------------------------------------------------
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