MUL operation result

Question

Please explain how MUL operation works in this situation:

MUL ECX

Before operation:

EAX: 000062F7 (25335) ECX: 3B9ACA00 (1000000000) EDX: 00000000 (0)

After operation:

EAX: C3ACE600 EDX: 0000170A (5898)

Would be thankful if anyone can explain me how 5898 was calculated.

Answer 1

Multiplication of two 32-bit values can yield a 64-bit result.

The result of MUL is stored in EDX:EAX. (The upper 32 bits in EDX, the lower 32 bits in EAX).

0x3b9aca00 * 0x62f7 = 0x170a c3ace600
   ECX        EAX       EDX    EAX

Answer 2

It's quite simple actually. The description of MUL r/m32 ( r/m32 refers to ECX in this case) is:

MUL r/m32  (EDX:EAX ← EAX ∗ r/m32).

So EAX is multiplied by ECX to form a 64-bit product which is stored in EDX:EAX (ie the low 32 bits end up in EAX , and the high 32 bits in EDX ).

So if we punch in 000062F7 * 3B9ACA00 in a calculator, we get 170AC3ACE600 , meaning that EAX will be C3ACE600 and EDX will be 170A (or 5898 in base10).