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I have used unions earlier comfortably; today I was alarmed when I read this post and came to know that this code union ARGB { uint32_t colour; struct componentsTag { uint8_t b; uint8_t g; uint8_t r; uint8_t a; } components; } pixel; pixel.colour = 0xff040201; // ARGB::colour is the active member from now on // somewhere down the line, without any edit to pixel if(pixel.components.a) // accessing the non-active member ARGB::components is actually undefined behaviour I.e. reading from a member of the union other than the one recently written to leads to undefined behaviour. If this isn't the intended usage of unions, what is? Can some one please explain it elaborately? Update: I wanted to clarify a few things in hindsight. The answer to the question isn't the same for C and C++; my ignorant younger self tagged it as both C and C++. After scouring through C++11's standard I couldn't conclusively say that it calls out accessing/inspecting a non-active union member is undefined/unspecified/implementation-defined. All I could find was §9.5/1: If a standard-layout union contains several standard-layout structs that share a common initial sequence, and if an object of this standard-layout union type contains one of the standard-layout structs, it is permitted to inspect the common initial sequence of any of standard-layout struct members. §9.2/19: Two standard-layout structs share a common initial sequence if corresponding members have layout-compatible types and either neither member is a bit-field or both are bit-fields with the same width for a sequence of one or more initial members. While in C, ( C99 TC3

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