Can anyone pleeaaasssseee tell me how does the compiler arrive at the output of this program. My predicted answer never matches with the actual output of this program.
Please try this out on paper first before compiling. I need a detailed explanation- step by step. Thanks....Code:#include <stdio.h> int main() { int n=3; printf("%d %d %d %d" , n, n++, ++n, n); //big confusion not able to get this one return 0; }
That is because what you are doing is undefined behaviour, so pretty much anything can happen. read here for a more thorough explanation: c++ - Undefined Behavior and Sequence Points - Stack Overflow
The code modifies n twice in one statement. According to all versions of the C and C++ standards, that results in undefined behaviour. When something is undefined according to the standard, any result is permitted. That result might be printing some value 4 times. That result may be the output of a set of distinct values that make sense to you. The result is permitted to be a set of values that don't make sense to you. The result is permitted to be reformatting your hard drive. If your compiler produced one result, and some other compiler produces a different result, then BOTH compilers are still correct. Whatever any compiler does is correct, because there are no defined constraints on what it is permitted to do when code exhibits undefined behaviour.
Actually I'm pretty sure that the program is implementation defined, not undefined, because the comma is a sequence point. Unfortunately, where the implementation defined part comes from is because the compiler can evaluate the arguments of the function printf() in any order.
Futher reading here: Comma Operator - C and C++ | Dream.In.Code
Last edited by whiteflags; 08-18-2012 at 03:22 PM. Reason: new better link
The commas that separate function arguments are not sequence points. The comma operator is different, even though it is the same symbol. This statement is defined:
But this is not:Code:a++, a++;
Code:f(a++, a++);
Well I guess the comma in the second place is just a token then. Oh well, my point doesn't change.
Incorrect. The comma in function calls is not the comma operator.Originally Posted by whiteflags
Also, C++-11 has removed references to "sequence points". The language now used in that standard reflects whether one operation is sequenced before another or not. The "or not" is explicitly called undefined behaviour now.
Regardless of version of C or C++ standard, however, the behaviour is undefined, not implementation defined.
Oh man, if I had a dollar for every time this question was asked...
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