C++ Question
What happens if the result of an arithmetic operation isn't stored in memory
When I was studying C++ 5 years ago, one of our assignments was
Create a program that calculates the temperature in fahrenheit based on the celsius input using the formula C° x 9/5 + 32 = F°
Our first version was something like this
int main()
{
float celsius;
cout << "Enter Celsius temperature: ";
cin >> celsius;
cout << "Fahrenheit: " << celsius * (9.0 / 5) + 32 << endl;
return 0;
}
A classmates pointed out that we weren't explicitly told to output the result, which resulted in
int main()
{
float celsius;
cout << "Enter Celsius temperature: ";
cin >> celsius;
celsius * (9.0 / 5) + 32;
return 0;
}
I've used this as an anecdote: always be thorough when specifying requirements
Lately I've been wondering if this code actually did satisfy the requirements.
Wouldn't the
celsius * (9.0 / 5) + 32;Viewing the Visual Studio Disassembly the statement doesn't seem to generate any code, but then again, neither does the
float celsius;C++ code and corresponding Disassembly
7: float celsius;
8: cout << "Enter Celsius temperature: ";
push offset string "Enter Celsius temperature: " (01368B30h)
mov eax,dword ptr [_imp_?cout@std@@3V?$basic_ostream@DU?$char_traits@D@std@@@1@A (0136C098h)]
push eax
call std::operator<<<std::char_traits<char> > (01361370h)
add esp,8
9: cin >> celsius;
mov esi,esp
lea eax,[celsius]
push eax
mov ecx,dword ptr [_imp_?cin@std@@3V?$basic_istream@DU?$char_traits@D@std@@@1@A (0136C09Ch)]
call dword ptr [__imp_std::basic_istream<char,std::char_traits<char> >::operator>> (0136C0A0h)]
cmp esi,esp
call __RTC_CheckEsp (0136114Fh)
10: celsius * (9.0 / 5) + 32;
11: return 0;
xor eax,eax
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Answer Source
Yes, looks like the compiler optimized the statement away. I bet if you used volatile float celsius; you would see the code!
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