Hello
The output of code1 gives any garbage value whereas output of code2 gives run time error.Why is this so?why both of the codes not showing an error because there are no arguments for %d & %f.
Code:/*CODE1*/ #include<stdio.h> #include<conio.h> int main(void) { clrscr(); printf("%d"); getch(); return 0; }ThanksCode:/*CODE2*/ #include<stdio.h> #include<conio.h> int main(void) { clrscr(); printf("%f"); getch(); return 0; }
It's undefined, so anything can happen. It can crash, it won't crash, it can eat up your computer to burn up your house. Anything can happen.
Originally Posted by Adak
As they say, this could fire a nuclear missile but that would be unlikely unless you have required hardware...
I might be wrong.
Quoted more than 1000 times (I hope).Thank you, anon. You sure know how to recognize different types of trees from quite a long way away.
And of course, if you have a good compiler, such as gcc, with warnings enabled, it will actually tell you that you are missing the arguments for printf, which is very handy when you are as bad at writing correct code as I am...
--
Mats
Compilers can produce warnings - make the compiler programmers happy: Use them!
Please don't PM me for help - and no, I don't do help over instant messengers.
I would add to the undefined behavior argument of the previous posts that printf() belongs to this particular set of functions taking a variable number of arguments which means the compiler has to do an extra effort to check whether the call is valid or not and this goes beyond the simple compilation of C as the its requires to know the semantic of the parameters (in the case of printf and scanf and equivalents, parse the format string, warnings are enabled for gcc with -Wformat[=2], this won't work for functions not using the standard 'format' parameter)
