C++17
Could someone explain how
int number{5};
number = (number++) + 10;
Gives an output of 15 while
int number {5};
number = (++number) + 10;
Gives an output of 16?
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Lightness Races in Orbit
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Andrew Mbugua
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6Why do you want to write code like this? – Lightness Races in Orbit Jan 07 '20 at 18:06
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1@LightnessRacesBY-SA3.0 - Maybe he does't want to write code like this, maybe he just wants to learn how to read code like this when he encounters it. – selbie Jan 07 '20 at 18:10
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1@selbie Maybe! Good thing I answered the question ;) – Lightness Races in Orbit Jan 07 '20 at 18:10
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@GuillaumeRacicot This is about more than just the difference between prefix and postfix. Sort of. – Lightness Races in Orbit Jan 07 '20 at 18:10
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@LightnessRacesBY-SA3.0 my bad. I'll vote reopen then. – Guillaume Racicot Jan 07 '20 at 18:11
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@GuillaumeRacicot I'm ambivalent tbh. On the one hand we don't really need yet another of these questions. On the other, meh – Lightness Races in Orbit Jan 07 '20 at 18:11
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Thankyou for your answers,sorry I could'nt respond sooner – Andrew Mbugua Jan 17 '20 at 19:09
3 Answers
9
Before P0145 was adopted (in C++17), the first example had undefined behaviour. Anything could happen.
Before C++11, both had undefined behaviour.
In C++17, neither has undefined behaviour. That doesn't mean it's code you want to be writing.
The explanation of your output is simple, if we understand the difference between postfix and prefix increment:
Case 1
number++:numberbecomes 6 but the expression evaluates to 5- ten is added to the expression
- the result (15) is stored in
number
Case 2
++number:numberbecomes 6 and the expression evaluates to 6- ten is added to the expression
- the result (16) is stored in
number
Lightness Races in Orbit
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1The pre-increment version has been well-defined since C++11, since the side-effect of the increment is sequenced-before its value computation which is sequenced-before the assignment's side-effect. – walnut Jan 07 '20 at 18:22
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@walnut Fair, I was considering the whole question as one program – Lightness Races in Orbit Jan 07 '20 at 18:32
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Pre increment increments the value and returns the incremented result. Post increment returns the value before the increment, then increments the variable.
Jesper Juhl
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Unary post-increment is executed after the statement executed. so, the number is added first and then number is 15. But in pre-increment case the unary operator excited first then the current statement is resolved and result is calculated to 16.
More information: https://en.cppreference.com/w/cpp/language/operator_incdec
int number{ 5 };
01013C88 mov dword ptr [number],5
number = (number++) + 10;
01013C8F mov eax,dword ptr [number]
01013C92 add eax,0Ah
01013C95 mov dword ptr [number],eax
01013C98 mov ecx,dword ptr [number]
01013C9B add ecx,1
01013C9E mov dword ptr [number],ecx
Assmebly code show the two time addition
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this has nothing to do with when "the statement executed". The link you should be giving is https://en.cppreference.com/w/cpp/language/eval_order – Cubbi Jan 07 '20 at 19:55
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int number{ 5 }; 01013C88 mov dword ptr [number],5 number = (number++) + 10; 01013C8F mov eax,dword ptr [number] 01013C92 add eax,0Ah 01013C95 mov dword ptr [number],eax 01013C98 mov ecx,dword ptr [number] 01013C9B add ecx,1 01013C9E mov dword ptr [number],ecx Assmebly code show the two time addition – Build Succeeded Jan 08 '20 at 05:18