Hi
I have a code:
In this assert macro will call only if x is 0.Code:#include<stdio.h> #define assert(cond) if(!(cond))\ printf("assertion failed for condition %s\n",#cond) main() { int x=100; if(x == 0) assert(x<100); else printf("No assert call\n"); getchar(); return 0; }
So normally as first if condition fail, it should come in else and print "No assert Call".
But it is not coming in else part and directly comes to getchar().
Can any body help why this happen?
Thanks
It works when you put brackets on the if/else:
If you don't want to have brackets, you should encapsulate your assert function in brackets (with a do-while).Code:if (x == 0) { assert(x<100); } else { printf("No assert call\n"); }
Your else in code "belongs" to the if in the macro
Code:if (x == 0) if (x < 100) printf(...); else printf("No assert call\n"); getchar();
Normally, when you write multi-line macros, you express them as a while loop.
When used in something like thisCode:#define assert(cond) do { \ if(!(cond)) \ printf("assertion failed for condition %s\n",#cond); \ } while ( 0 ) /* trailing ; is deliberately omitted */
it has the effect of making it seem to the outer level if statement that the assert is a single statement, and the else will bind to the obviously intended if (and NOT the if now buried inside a while loop).Code:if(x == 0) assert(x<100); else printf("No assert call\n");
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
