Solidity: using input of `0` in sha3 function produces incorrect output

[elenadimitrova]

I suspect this is a bug but wanted to report and confirm.
When passing 0 to sha3 function in Solidity, instead of a predefined uint zero-value variable the produced output is incorrect. Sample below

contract MyContract
{
    bytes32 public sha3test1;
    bytes32 public sha3test2;
    uint256 zero = 0;

    function MyContract(){

       sha3test1 = sha3(0);
       sha3test2 = sha3(zero);
    }
}

produces:

sha3test1 = 0xbc36789e7a1e281436464229828f817d6612f7b477d66591ff96a9e064bcc98a
sha3test2 = 0x290decd9548b62a8d60345a988386fc84ba6bc95484008f6362f93160ef3e563

[elenadimitrova]

@area has rightly pointed out that Solidity casts to the smallest type hence using a uint8 for the supplied 0 value. Tests confirm this is the case so closing this.

[axic]

I'm not sure that is the correct behaviour though for sha3.

In the above case you coerce with sha3(uint256(0));, but I would be include to say that constants for tight packing should either:

• be disallowed
• or taken as uin256.

If finer granularity is needed they can be typecasted or defined as variables.

[axic]

The document here could be made a tiny bit more explicit about how it behaves.

[beether]

Ok, so sha3(zero) doesn't work... how does one get it to work as expected? It would be quite helpful to show how to actually hash a f*ing uint256 somewhere. Tightly packed or not, I don't really care and just want it to work like any sane person would expect.

[chriseth]

@beether how do you expect it to work? sha3(0) should be the same as sha3(uint8(0)), should be the same as

uint 8 x = 0;
sha3(x);

On the other hand, sha3(uint(0)) is a different value which should be equal to uint x = 0; sha3(x);.