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It is undefined whether the left or right side of the addition is accessed first. Accesses to volatile memory locations can have side effects e.g. if it is an I/O device register and you don't want the compiler to decide for you what the order will be. Plus, the compiler or C++ standard probably can't assume the special case that there is only one thread.

Which website is that

>>25504646 but won't it have anyway the same output as 0? >>25504931 cppquiz.org

just dump it into chatgpt and ask

>>25505199 of course it answered incorrectly

retard

>>25505932 Nigger

hmm i always thought using volatile doesn't create UB just a keyword to put a variable into special quick memory

Let’s break down this C++ program and the explanation provided, focusing on the behavior according to the C++23 standard and why the output is considered undefined. The Program cpp #include <iostream> volatile int a; int main() { std::cout << (a + a); } This is a simple C++ program that: Includes the <iostream> header for input/output operations. Declares a global variable a of type volatile int, which is implicitly initialized to 0 (since it has static storage duration and no explicit initializer). In main(), it outputs the result of a + a using std::cout. At first glance, you might expect the program to output 0 because a is implicitly initialized to 0, and 0 + 0 = 0. However, the explanation highlights a critical issue related to the volatile keyword and the C++23 standard’s rules about sequencing and side effects. Key Concepts from the Explanation 1. The volatile Keyword The volatile qualifier indicates that the variable’s value can change unexpectedly (e.g., by hardware, another thread, or external factors). In C++, reading or writing a volatile object is considered a side effect. This means that accessing a (reading its value) is not just a simple value retrieval—it has observable behavior that the compiler must respect. 2. Undefined Behavior Due to Unsequenced Side Effects According to the C++ standard, if two side effects on the same memory location (like reading a) are unsequenced (i.e., the order of their evaluation is not guaranteed), and they are not explicitly ordered by the program, the behavior is undefined. In the expression a + a, there are two reads of the volatile variable a. The C++ standard does not guarantee the order in which these reads occur because the evaluations of the operands of the + operator are unsequenced relative to each other. Since both reads are side effects (because a is volatile), and they are unsequenced, this leads to undefined behavior. 3. Why Not Just 0? You might think a is 0 (due to implicit initialization) and a + a should be 0 + 0 = 0. However, the volatile nature of a means the compiler cannot assume a’s value is stable or predictable between reads. The undefined behavior arises from the lack of sequencing, not necessarily from the value of a. In a single-threaded program (as implied here, since no concurrency is mentioned), the issue is still present because the standard treats unsequenced side effects on the same memory location as undefined, regardless of threading. 4. Implications of Undefined Behavior Undefined behavior means the program’s output is not predictable and could produce any result, including crashing, outputting garbage, or appearing to work correctly on some compilers or platforms but not others. In practice, a compiler might optimize or reorder the reads of a in ways that lead to unexpected results, especially since volatile forces the compiler to avoid certain optimizations but doesn’t guarantee sequencing. Why This Is a Difficult Question (Difficulty: ★★★) The question requires understanding advanced C++ concepts like: The semantics of volatile and its role in side effects. The C++ memory model and rules for sequencing (or lack thereof) in expressions. The distinction between undefined behavior and implementation-defined behavior. Many programmers might overlook the volatile keyword’s impact or assume the expression a + a is straightforward, but the standard’s strict rules create a subtle trap. Conclusion According to the C++23 standard, the output of this program is undefined due to the unsequenced reads of the volatile variable a in the expression a + a. The program does not guarantee any specific output, and any observed behavior (including outputting 0) is not reliable across all compilers or platforms. If you’d like to explore this further or see how different compilers handle it, I could offer to look into real-world behavior, but based on the standard alone, the answer is that the output is undefined. Let me know if you’d like to dive deeper!

>>25509077 Faggot

I don't see what's undefined about this. Even if the order of reading the operands is unspecified, addition of ints is still commutative and the result will be the same. If you compile this, you will see that the only difference volatile makes is that the variable is accessed twice.

>>25509103 Let's address your questions one at a time, focusing on the behavior of the volatile keyword and the specific C++ program provided. 1. Does the volatile keyword put variables into a special quicker-to-access memory? Answer: No, the volatile keyword does not inherently place variables in faster memory. Its primary purpose is to prevent the compiler from optimizing away or reordering accesses to the variable, ensuring that reads and writes are treated as side effects. Explanation: What volatile Does: The volatile keyword tells the compiler that the variable's value can change unexpectedly, such as by hardware, another thread, or external factors. This means: Every read of a volatile variable must be performed as a memory access (side effect), not optimized away or cached in registers. Writes to a volatile variable must also be performed as memory accesses and not optimized or reordered. It ensures that the compiler treats accesses to the variable as visible side effects, which is crucial for scenarios like memory-mapped I/O or hardware registers. Memory Placement: The volatile keyword does not specify or imply any particular memory location or type (e.g., faster memory like registers or cache). It only affects how the compiler generates code for accesses to the variable, not where the variable is stored in memory. Variables marked volatile are typically stored in the same memory as other variables (e.g., RAM), unless explicitly placed elsewhere by linker scripts or hardware-specific configurations. Performance Implications: While volatile prevents certain optimizations (like caching values in registers), which might indirectly affect performance, it does not guarantee faster access. In fact, forcing memory accesses instead of using registers can sometimes make access slower, depending on the architecture. Why the Misconception? Some people confuse volatile with memory-mapped I/O or hardware registers, where variables might indeed be placed in special memory locations. However, volatile itself does not dictate memory placement; it only enforces that accesses are not optimized away or reordered. Conclusion: The volatile keyword is about ensuring that accesses to the variable are not optimized away or reordered, not about placing it in faster memory. Therefore, your initial thought that volatile puts variables into a "special quicker-to-access memory" is incorrect. 2. Analyzing the Possible Outcome Based on Likely ASM Code Given the program: cpp #include <iostream> volatile int a; int main() { std::cout << (a + a); } Answer: The assembly code will likely show two separate reads from the memory location of a, but due to the undefined behavior in the expression a + a, the exact sequence and outcome are unpredictable and compiler-dependent. Explanation: Undefined Behavior Recap: The expression a + a involves two unsequenced reads of the volatile variable a. According to the C++23 standard, reading a volatile object is considered a side effect (see [intro.execution]/7 and [basic.lval]/). Since the two side-effecting reads of a are unsequenced relative to each other, this leads to undefined behavior (see [intro.execution]/10 and [intro.memory]/). Undefined behavior means the compiler is not required to generate code that behaves in any specific way, and the output is unpredictable. Likely ASM Code: Because a is volatile, the compiler will likely generate two separate load instructions to read the value of a from memory. For example, on x86 assembly, this might look like: assembly mov eax, [a] ; Load a into eax mov ebx, [a] ; Load a into ebx add eax, ebx ; Add ebx to eax However, due to undefined behavior, the compiler might interleave other instructions or optimize in unexpected ways, especially since the reads are unsequenced. Possible Outcomes: If a doesn't change: In a single-threaded program where a isn't modified externally, both reads might retrieve the same value (e.g., 0, since a has static storage duration and is implicitly initialized to zero). The output could be 0, but this is not guaranteed due to undefined behavior. If a changes externally: If a is modified by hardware or another process between the two reads, the two reads could get different values, leading to an unpredictable sum. Compiler-Specific Behavior: Different compilers might handle the unsequenced reads differently: One compiler might read a twice in quick succession. Another might cache one read in a register (though less likely due to volatile). Some compilers might even optimize the expression in unexpected ways, given the undefined behavior. Why It's Undefined: The C++ standard does not specify the order of evaluation for the operands of +, so the two reads of a could happen in any order or even simultaneously (though simultaneous reads aren't typically possible in assembly). Since both reads are side effects and are unsequenced, the standard deems this undefined behavior to allow compilers flexibility in optimization. Practical Advice: To avoid undefined behavior, ensure that side effects on the same memory location are sequenced. For example, you could write: cpp int temp = a; // Read a once std::cout << (temp + temp); // Use the temporary This guarantees that a is read only once, and the expression uses that single value. Conclusion: While the likely assembly code will show two separate reads of a, the undefined behavior means that the actual outcome is unpredictable and could vary between compilers or even different runs on the same compiler. The output is not guaranteed to be consistent, and relying on it is unsafe. If you'd like, I can look into how specific compilers (like GCC or Clang) handle this code and provide more concrete examples of the generated assembly. Let me know if that would be helpful!

according to grok, casting volatile to int like this #include <iostream> volatile int a; int main() { std::cout << ((int)a + (int)a); } wouldn't help. however casting normal int to volatile makes sure its side effecte'd

From what I understand, unless a is actually modified this piece of code will always print 0.

>>25509168 you're supposed to say it's UB even if compilers are deterministic just because c++ people suck cock at standardization conferences

>>25509172 Okay got it

If you read the part of the standard they're citing, you'll see an example that actually makes sense:

void g(int i) {
  i = 7, i++, i++;              // i becomes 9

  i = i++ + 1;                  // the value of i is incremented
  i = i++ + i;                  // undefined behavior
  i = i + 1;                    // the value of i is incremented
}

Here the outcome depends on which operand is evaluated first. In their example, it does not.

>>25509184 My face when they ask this in an interview. Like seriously, nobody codes like that. Why would I know the answer? t. been coding C++ for over decade

Nobody ever needed a volatile variable.

>>25515609 They are needed in embedded. But they're not used like this.

>>25515609 Widely used in multithreading code, OS kernels, hacking, and more, my dear low IQ Bernd.

>>25515707 >multithreading code Yes, but only if you are using it wrong. >OS kernels Grep'd linux sources, volatile was used three times. C++ was used: Never. >hacking Bro you just posted cringe.