# How to correctly multiply signed int with unsigned int and store the result in `uint32`?

I have an `int32` val that contains some negative value such as `-1`. I want to multiply it with a `uint32` value and add a larger `uint32` value (for which I know the result will be positive) and store in `uint32`.

I have come up with the following solution:

``````uint32 result = uint32( uint32(i) * val + 100)
``````

For example: `i=-1`, `val=10`, result should be (-1)*(10) + 100 = 90.

[Q] Is this the correct way to do it?

Since I am casting a `signed int` to an `unsigned int` as ( `uint32(i)` ), shouldn't a negative value (-1) get automatically converted into positive (+1) and give a result of (+1)*(10) + 100 = 110?

`````` pragma solidity ^0.4.6;
contract NumTest {
int8 i;

function NumTest() {
i = -1;
}

function number() constant returns (uint32 num) {
uint32 val = 3;
return uint32( uint32(i) * val + 20);
}
}
``````

Just commenting on this part...

shouldn't a negative value (-1) get automatically converted into positive (+1)

No, because integers are represented as Two's Complement.

See this previous post for a detailed explanation: Math operation between int and uint

Casting `-1` with `uint32(-1)` gives you the value 4294967295 (`0b11111111111111111111111111111111` or `0xFFFFFFFF`). This is the maximum value an unsigned 32-bit integer can take. This is equivalent to a signed integer value of `-1` because of the integer being represented in Two's Complement, as mentioned above.

`uint32(i) * val` respectively gives you two less: 4294967293 (`0b11111111111111111111111111111101` or `0xFFFFFFFD`).

If we then add 20, we overflow the (unsigned) integer, which gives us 17. So it looks like it's working by adding 20 to -3, but actually it's adding 20 to 4294967293.

• As I understand, I am good to cast negative value(-1) into uint32 on do the equation I mentioned without having any unexpected value? @RichardHorrocks – alper Feb 22 '17 at 16:31
• Yes, overall it will work :) But not by changing `-1` to `+1` directly. – Richard Horrocks Feb 22 '17 at 17:50

From other languages, you can expect that `uint(-1)== 1`, but this is not the case with solidity.
From documentation:

``````uint8 x = -3;
unit x = uint(y);
``````

Here at end of code snippet, x will have value -3.