Assembly: arithmetic operators?

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zutokaza

Joined: 03 May 2015
Posts: 14

zutokaza 06 May 2015, 09:45

How would this be done in windows assembly? (32 bit)

Code:

int main()
{
     cout << "7 + 3 = " << 7 + 3 << endl;
     cout << "7 - 3 = " << 7 - 3 << endl;
     cout << "7 * 3 = " << 7 * 3 << endl;
     cout << "7 / 3 = " << 7 / 3 << endl;
     cout << "7.0 / 3.0 = " << 7.0 / 3.0 << endl;
     cout << "7 % 3 = " << 7 % 3 << endl;
     cout << "7 + 3 * 5 = " << 7 + 3 * 5 << endl;
     cout << "(7 + 3) * 5 = " << (7 + 3) * 5 << endl;
     return 0;
}

06 May 2015, 09:45

revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 19869
Location: In your JS exploiting you and your system

revolution 06 May 2015, 12:56

X86 instructions have ADD, SUB, MUL and DIV as explained in the manuals. DIV also gives the remainder so it can be used for modulus.

06 May 2015, 12:56

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