I’m trying to grasp the Cvalue idea in MISRA C++ 2008. The definition is as follows:
“An expression that should not undergo further conversions, either implicitly or explicitly, is called a cvalue expression.”
Nevertheless, the provided code examples in the rule show:
int32_result = static_cast<int32_t>(byte_val) + byte_val; // Example 1 - Valid
int32_result = int32_result + byte_val; // Example 2 - Valid
It looks like the operands on the right side of these addition operations are experiencing both implicit and explicit conversions. Despite the rule stating that these examples are valid, this appears to be in conflict with the Cvalue definition. Can anyone clarify how these examples fit with the Cvalue concept? I’m puzzled about why these conversions occur if Cvalue expressions shouldn’t experience any further conversions.
You’re reading “should not undergo further conversions” too literally. A cvalue expression is just the final result you want from your calculation at that point. Here’s what MISRA actually means - once your expression gets evaluated to its intended type, that’s when no more conversions should happen. In your example static_cast<int32_t>(byte_val) + byte_val, the conversions you see are just normal arithmetic rules doing their thing. The byte_val gets promoted to int32_t to match the cast, then you get addition. That resulting int32_t expression is your cvalue - it’s what you wanted to compute. The rule just stops this cvalue from getting accidentally converted to something else later. It’s not trying to kill conversions during evaluation - it’s making sure your final computed value keeps its expected type without surprise conversions that could cause bugs or lose precision.
you’re overcomplicating this. cvalue just means “this expression has reached its final form.” the conversions in your examples happen before the cvalue stage, not after. so when you do static_cast<int32_t>(byte_val) + byte_val, all the casting and promotion happens first, then you get your int32_t result - that’s the cvalue. misra just wants to prevent that final result from getting converted again unexpectedly.
You’re mixing up what a “cvalue expression” actually means here. A cvalue isn’t about stopping all conversions - it’s about spotting expressions that show the final value type you want for an operation. In your examples, conversions happen during evaluation, not to the cvalue expressions themselves. When you write static_cast<int32_t>(byte_val) + byte_val, the cast makes an int32_t value and the second operand gets promoted to match. The whole expression becomes the cvalue - that final computed value that shouldn’t get hit with more unwanted conversions. This rule prevents situations where you do arithmetic thinking you’re using one type, but the result gets converted to something completely different without warning. Your examples work fine because the conversions are either explicit (the cast) or expected promotions that don’t mess with what the operation’s supposed to do.