if (n!=0) n=0; v/s n=0; Which is more efficient and why?

Question

While typing a program as a high level programmer, n = 0; looks more efficient and clean.

But is n = 0; really more efficient than if (n != 0) n = 0;?

1. when n is more likely to be 0.

2. when n is less likely to be 0.

3. when n is absolutely uncertainty.

Language: C (C90)

Compiler: Borland's Turbo C++

Minimal reproducible code

void scanf();

void main()
{
int n; // 2 bytes

n=0; // Expression 1

scanf("%d",&n); // Absolutely uncertain

if(n!=0) n=0; // Expression 2

}

Note: I have mentioned the above code only for your reference. Please don't go with it's flow.

If your not comfortable with the above language/standard/compiler, then please feel free to explain the above 3 cases in your preferred language/standard/compiler.

Answer 1

If n is a 2's complement integral type or an unsigned integral type, then writing n = 0 directly will certainly be no slower than the version with the condition check, and a good optimising compiler will generate the same code. Some compilers compile assignment to zero as XOR'ing a register value with itself, which is a single instruction.

If n is a floating point type, a 1s' complement integral type, or a signed magnitude integral type, then the two code snippets differ in behaviour. E.g. if n is signed negative zero for example. (Acknowledge @chqrlie.) Also if n is a pointer on a system than has multiple null pointers representations, then if (n != 0) n = 0; will not assign n, when n is one of the various null pointers. n = 0; imparts a different functionality.

"will always be more efficient" is not true. Should reading n have a low cost, writing n a high cost (Think of re-writing non-volatile memory that needs to re-write a page) and is likely n == 0, then n = 0; is slower, less efficient than if (n != 0) n = 0;.

Answer 2

n = 0;

will always be more efficient as there is no condition check.

https://godbolt.org/z/GEzfcD