Hi,
I have seen in embedded prgarmming many of using (void)printf(),(void)scanf()......like this .
printf is a function which returns an integer that is many character has been written but if we r not intersted in the return value then it is lossed.
but when we r specifying void infront of printf is that compiler ignores that return value?.......
Is ther any performance related thing with this ? I have tested time concern thing but I didn't find anything.
could anybody help me in this concept.
6  2187 
Not completely sure what that's doing. I've always used cin and cout rather than printf and scanf...Perhaps it is taking the integer that was returned by both and typecasting it to type void - in effect, deallocating that integer from memory.
Its type casting an integer return value to void, which basically means its nullifying the return value. AFAIK, it doesn't garner any performance gains. If anything the casting takes up more flops then just returning the int to nothingness.
The reason for the cast to void is to keep the compiler from complaining. There are some compilers which are very strict. These compilers will take offence at a function call return value not being used, and assumes that you did something that you didn't intend to do. Casting it to void tells the compiler that you indeed intend to ignore the return value, and so will no longer complain.
More and more, in new code, people are not allowing warnings to just happen. The reason for this is because warnings are there to WARN the programmer that they may have done something incorrect. Telling the compiler that you intended to do what you did keeps the compiler happy and allows the programmer not to get so flooded with warnings that they are ignored, possibly allowing a programmer to do something unintentional.
Oh and casting to void takes no flops. ;) In fact there are many casts that are only cosmetic and have no actual code generated from them. Doing an upcast (casting to a derived class) on a single inheritance type for instance does nothing but tell the compiler to think of the object as its more general form. It too doesn't actually generate any code. Reinterpret casts to a reference or a pointer is also of that class of casts.
Hope this helps.
Adrian
Great post! Thanks Adrian!
The reason for the cast to void is to keep the compiler from complaining. There are some compilers which are very strict. These compilers will take offence at a function call return value not being used, and assumes that you did something that you didn't intend to do. Casting it to void tells the compiler that you indeed intend to ignore the return value, and so will no longer complain.
More and more, in new code, people are not allowing warnings to just happen. The reason for this is because warnings are there to WARN the programmer that they may have done something incorrect. Telling the compiler that you intended to do what you did keeps the compiler happy and allows the programmer not to get so flooded with warnings that they are ignored, possibly allowing a programmer to do something unintentional.
Oh and casting to void takes no flops. ;) In fact there are many casts that are only cosmetic and have no actual code generated from them. Doing an upcast (casting to a derived class) on a single inheritance type for instance does nothing but tell the compiler to think of the object as its more general form. It too doesn't actually generate any code. Reinterpret casts to a reference or a pointer is also of that class of casts.
Hope this helps.
Adrian
Heh, who would have known its just a compiler thing. I thought it was just a programmer choice, you know some programmer's like to be extremely explicit about things. Now that I think about it I see how there is not code generated from that type of cast. Interesting!
THANKS AdrianH,
Ur information help me lot.
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