is there some way to elaborate this line so that it works
Code://------------------------------------------------------------------------------ int main() { int aaa ; for (aaa=0 ; aaa=3 , aaa=6 ,aaa=9 ;aaa+++) { printf("\n %d \n", aaa); } return 0; } //----------------------------------------------------------------------------
Oh... there is ONE error: The expression a+++ is wrong. Everything else "works".
Originally Posted by flp1969
.Code:int main() { int aaa ; for (; aaa=3 , aaa=6 , aaa=9 ;) { printf("\n %d \n", aaa); } return 0; } .
strange to me print 9 alone boh!!!
i mean something like that
Code:[int main() { int aaa ; for (aaa=0 ; aaa<=9 ; aaa++ ) if (aaa==3 || aaa==6 || aaa ==9) { printf("\n %d \n", aaa); } return 0; }
Maybe
for ( aaa = 3 ; aaa <= 9 ; aaa += 3 )
Cut out all the excess.
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.
I said it works, not if it's doing what you want...
The for statement works with 3 arguments: initialization, condition to remain in loop and update:
.Code:for (init; cond; upd) { ... }
It is a simplified way to write:
What you are doing is:Code:init; while (cond) { ... upd; }
Since the comma operator will evaluate the sub-expressions from left to right, 'aaa=3, aaa=6' will be discarded. Note the '=' operator. When you do an assignment, the resulting value (9) is considered as true, so the loop never ends, and aaa will be always 9.Code:// for (; aaa = 3, aaa = 6, aaa = 9;) ... while ( aaa=9 ) { printf( " %d\n", aaa ); }
It is not "strange". It is what you said the compiler to do.
Last edited by flp1969; 03-18-2019 at 01:07 PM.
