I came across the following puzzle somewhere
#include <stdio.h> int main() { { /*Fill in something here to make this code compile ........... */ ooOoO+=a; } #undef ooOoO printf("%d",ooOoO); return 0; } In short I want to ask how can I use ooOoO in printf after it has been #undef ed?
You need to declare it as a variable:
#define ooOoO int ooOoO = 42; int a = 1; { ooOoO Macro-replacement is non-recursive; while ooOoO is being replaced, the identifier ooOoO will not be treated as a macro name.
If you are looking for a solution that does not use a macro, then you can simply ignore the #undef directive and never declare ooOoO as a macro. It is permitted in both C and C++ to #undef an identifier that is not defined as a macro.
main() was a typo; if it isn't, then the block it introduces would need to be closed anyway before declaring ooOoO as a variable, otherwise it would not be in scope when the printf statement is reached.After reformatting the code (indent) and adding the solution, that's what I receive:
#include <stdio.h> int main() { { /*-Insert starts here-*/ } int ooOoO = 0, a=3; { /*-Insert ends here-*/ ooOoO+=a; } #undef ooOoO printf("%d",ooOoO); return 0; } compiles and prints 3
How about this?
#include <stdio.h> int main(){ int ooOoO = 0; { int a = 3; ooOoO+=a; } #undef ooOoO printf("%d",ooOoO); return 0; } 
#include <stdio.h> int main(){ { /*Fill in something here to make this code compile */ } int a = 0, ooOoO=0; #define ooOoO ooOoO { /* */ ooOoO+=a; } #undef ooOoO printf("%d",ooOoO); return 0; } 
} missing at the end and therefore doesn't compile.The #undef undefines the symbol to the preprocessor so that it does not get substituted with something else, but ooOoO still gets to the compiler.
main(); is that intended?