Question

I have the following code in C:

        int l;
        short s;

        l = 0xdeadbeef;
        s = l;

Assuming int is 32 bits and short is 16 bits, when performing s = l, s will be promoted to 32 bits and after assignment, only lower 16 bits will be kept in s. My question is that when s is promoted to 32 bits, will the additional 16 bits be set to 0x0 or 0xf ?

Source : http://www.phrack.com/issues.html?issue=60&id=10

Answer 1

Actually s is not promoted at all. Since s is signed and l is too large to fit in s, assigning l to s in this case is implementation defined behavior.

6.3.1.3-3

Otherwise, the new type is signed and the value cannot be represented
in it; either the result is implementation-defined or an
implementation-defined signal is raised.