C Language, casting printf/scanf

casting printf/scanf

Author	Message
Jos A. Horsmeie #1 / 8	casting printf/scanf Quote: > Another question from a beginner... Is there any difference > between printf( ... ) and (void)printf( ... )? Same for scanf. > Which form should I use? Thanks. FC. The second form takes care of (some) compilers that can't keep their pedantic mouth shut. The printf() function returns a value (the number of characters actually written or a negative value if something went wrong). A simple statement like 'printf("Hello world\n");' discards this return value implicitely, while the second form '(void)printf("Hello world\n");' explicitely tells the compiler that you're not interested in printf's return value. Both forms are perfectly legal and both generate identical code. Ignoring the return value of a function call is AVND (A Very {filter} Deed) by itself, but I do prefer the first form because a simple function call, not part of another expression (including assignment expressions) tells me that I'm simply calling this function for its side effects (identical to Pascal's 'procedure' vs. 'function' calls). If a compiler warns about it, I simply consider it a 'nanny-type' of warning after checking out what actually made the compiler whine ... kind regards,
Sat, 04 Mar 2000 03:00:00 GMT

Francisco Cribar #2 / 8	casting printf/scanf Another question from a beginner... Is there any difference between printf( ... ) and (void)printf( ... )? Same for scanf. Which form should I use? Thanks. FC.
Sat, 04 Mar 2000 03:00:00 GMT

Craig Franc #3 / 8	casting printf/scanf Quote: >Another question from a beginner... Is there any difference >between printf( ... ) and (void)printf( ... )? Same for scanf. printf(...) ignors the functions retrun value whereas the void cast tells the compiler you don't want to know what it was. Quote: >Which form should I use? Thanks. FC. Some compilers and source code checkers may generate a warning if you throw away the value returned by a function. With printf, what you are primarily interested in is the what the function does, and not what it returns (the number of characters transmitted, or a negative value if an output error occured). Most people don't care about the first part, and I'm not sure what the proper response to a failure on a call to printf would be. I think all those void casts just clutter up the code. Note that you are still ignoring a possible failure condition. See if you can configure your compiler or lint pakage so as not to generate useless warnings. -- Craig Manchester, NH I think the military is having a real problem at the moment -- it has to train its recruits to kill, but not to offend. -- Arianna Huffington
Sat, 04 Mar 2000 03:00:00 GMT

Mark S. Novojilo #4 / 8	casting printf/scanf Quote: > Another question from a beginner... Is there any difference > between printf( ... ) and (void)printf( ... )? Same for scanf. > Which form should I use? Thanks. FC. Compiler never differ functions from their return value, only from their names and (in C++)argument types (overloading). That means printf( ... ) and (void)printf( ... ) is just the same. - Mark
Sun, 05 Mar 2000 03:00:00 GMT

Craig Franc #5 / 8	casting printf/scanf Quote: >> Another question from a beginner... Is there any difference >> between printf( ... ) and (void)printf( ... )? Same for scanf. >> Which form should I use? Thanks. FC. >Compiler never differ functions from their return value, only from >their names and (in C++)argument types (overloading). >That means printf( ... ) and (void)printf( ... ) is just the same. They represent two totally different kinds of expressions. A void expression denotes a nonexistent value. That is what the cast causes the value returned from printf to become. A plain printf call may result in 17, or 32, ect. -- Craig Manchester, NH I think the military is having a real problem at the moment -- it has to train its recruits to kill, but not to offend. -- Arianna Huffington
Mon, 06 Mar 2000 03:00:00 GMT

Pete Becke #6 / 8	casting printf/scanf Quote: > >> Another question from a beginner... Is there any difference > >> between printf( ... ) and (void)printf( ... )? Same for scanf. > >> Which form should I use? Thanks. FC. > >Compiler never differ functions from their return value, only from > >their names and (in C++)argument types (overloading). > >That means printf( ... ) and (void)printf( ... ) is just the same. > They represent two totally different kinds of expressions. A void > expression denotes a nonexistent value. That is what the cast causes > the value returned from printf to become. A plain printf call may > result in 17, or 32, ect. At the risk of sounding overly picky: the return type of printf is int, regardless of what casts you may apply to it. The result of a cast to void is void, so the type of the expression (void)printf(...) is void, but printf still returns its integer value. The cast just says to ignore it. -- Pete
Mon, 06 Mar 2000 03:00:00 GMT

Mark Brow #7 / 8	casting printf/scanf Quote: > Another question from a beginner... Is there any difference > between printf( ... ) and (void)printf( ... )? Same for scanf. printf() returns a value (the number of characters written or an negative number to indicate an error). Simply calling printf() as in your first example discards this return value. Casting its return value to void tells the compiler that yes, you really do intend to ignore the result. Some compilers and lints will warn you you are ignoring the return value if you don't do this. A similar thing happens with scanf() except that in this case it is meaningful to check the return value because it tells you the number of fields successfully read in. If you must use scanf() for input, then you should make every effort to make sure you handle all the input correctly. Quote: > Which form should I use? Thanks. FC. For printf(), it's a matter of taste. Most people find the cast overly picky, and few compilers or lints will warn about it's omission. For scanf() the result can be used meaningfully, so it shouldn't be being ignored. -- http://www.geocities.com/SiliconValley/Lakes/7537/
Wed, 08 Mar 2000 03:00:00 GMT

Craig Franc #8 / 8	casting printf/scanf Quote: >> >> Another question from a beginner... Is there any difference >> >> between printf( ... ) and (void)printf( ... )? Same for scanf. >> >> Which form should I use? Thanks. FC. >> >Compiler never differ functions from their return value, only from >> >their names and (in C++)argument types (overloading). >> >That means printf( ... ) and (void)printf( ... ) is just the same. >> They represent two totally different kinds of expressions. A void >> expression denotes a nonexistent value. That is what the cast causes >> the value returned from printf to become. A plain printf call may >> result in 17, or 32, ect. >At the risk of sounding overly picky: the return type of printf is int, >regardless of what casts you may apply to it. The result of a cast to >void is void, so the type of the expression (void)printf(...) is void, >but printf still returns its integer value. The cast just says to ignore >it. It would have been clearer if I had written "the value of the expression", rather than "the value returned from printf". If printf returned 17, you would have (void)17. Interestingly, Mark's comment about printf(...) and (void)printf(...) being the same, with a slight shift in context (a'la SN), has a grain of truth to it: They probably result in the same code being generated. It takes something like int i = printf(...) to get the returned value popped off the stack (you should probably expect this). The void cast is working at the C language level, not at the level of processor intstructions. -- Craig Manchester, NH Maybe a great magnet pulls/All souls towards truth Or maybe it is life itself/That feeds wisdom/To its youth. -- k.d. lang
Wed, 08 Mar 2000 03:00:00 GMT

Page 1 of 1

[ 8 post ]