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I happen to know of other methods used to swap two variables without a temporary variable like addition and subtraction, bitwise and so on. But I want to know why this method does not give correct result. The code is meant for C language.

int a=15, b=7;
a= b+(1 && (b=a)*0);
printf("%d %d",a, b);
return 0;

What I have tried:

Since && is a sequence point the first b should return 7 and second one should be assigned to 15?

So the output should be a=7 and b=15. But it is not.
Posted
Updated 18-May-21 19:13pm

Quote:
So the output should be a=7 and b=15. But it is not.

Obviously, your compiler use different rules than yours. And compiler rules may evolve with optimizations.
My robust swap methods are:
C++
int a=15, b=7;
int tmp= a; a= b; b=tmp; // and I let compiler optimizations remove the variable.

or
C++
int a=15, b=7;
a^=b; b^=a; a^= b; // swap by xor
   
Comments
Vishwash 2021 19-May-21 3:36am
   
So what does C standard tell about the given code? This appeared in a test where they want us to tell whether this will work or not.
Patrice T 19-May-21 5:26am
   
The only way to know for good is to look at assembly generated by compiler.
C++
a= b+(1 && (b=a)*0);
shouldn't that use the logical comparison, == for (b=a) ?

Also, why are you multiplying by 0? The effect is (1 && 0).
   
Comments
Patrice T 19-May-21 0:59am
   
"shouldn't that use the logical comparison, == for (b=a) ?"
No he wants initial value of a being assigned to new b.
"Also, why are you multiplying by 0? The effect is (1 && 0)."
it is a complicated way to isolate the assignment to b.

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