Expression “variable, variable = value;”

六眼飞鱼酱① 提交于 2019-12-01 18:44:13

This is likely an error in the program. The statement

a, b = c;

Is completely equivalent to

b = c;

Since the comma operator evaluates from left to right and discards all values except the last. Since the expression a has no side effects, it's essentially a no-op.

I would suspect that this is either programmer error or an incorrect translation of code from a different language into C++. You should contact the author to let them know about this.

Hope this helps!

Legal but questionable. The part before the comma doesn't do anything at all.

something, somethingElse = 0; 

probably, it is done to avoid the unused variable warning on variable something an to initialize the somethingElse variable to 0.

Does this make any sense in C++?

Yes syntactically it does, but without comments you may not know the developers intentions were (if any) other than maybe suppressing a variable warning.

Is a variable name also an expression?

Yes a variable itself is an expression. Ex. if(<expression>) if(something)

This code does compile, so how does this work?

It works by using the comma operator and ignoring the result of something then assigning 0 to somethingElse. Although something was marked volatile the original developer may of had a compiler that still complained about unused variables and being the clever developer he or she was then decided to suppress with that syntax.

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