does in c++ the conversion from unsigned int to int always preserve the bit pattern?

Question

From the standard (4.7) it looks like the conversion from int to unsigned int, when they both use the same number of bits, is purely conceptual:

If the de

Answer 1

int is the destination type in this case. As you say 2^32-1 cannot be represented so in this case so it is implementation-specific. Although, I've only ever seen it preserve bit patterns.

EDIT: I should add that in the embedded world often whats done when one storage location needs multiple representations that are bit-for-bit identical we often use unions.

so in this case

union FOO {
    int32_t signedVal;
    uint32_t unsignedVal;
} var;

var can be accessed as var.signedVal to get the 32 bits stored as a signed int and var.unsignedVal to get the 32 bits stored as an unsigned value. In this case bits will be preserved.