I am currently preparing for finals as we speak. I am unable to figure out the answer to this one and why it would be that answer. Would anybody be able to explain to me the answer and how it resulted in that? This is from an old exam. Thank you in advance!
int main(){
int i = 1, j=1, k=1;
printf("%d", ++i || ++j && ++k);
printf("%d %d %d \n", i, j, k);
return 0;
}
The program is a very simple example to understand the precedence, if you don't know the operator precedence look here. From my understanding this is what I think, let's take it step by step(following left to right association):
++ wins over everything, i is increamented:
int main(){
int i = 1, j=1, k=1;
printf("%d\n", ++i );
printf("%d %d %d \n", i, j, k);
return 0;
}
Output:
2
2 1 1
Here comes the logical operator, now you need to know in case of some expression like ( expr1 || expr2), expr2 gets executed when expr1 is zero or false, here i=2 hence it won't. Also, first printf prints true(1) or false(0) after executing ( expr1 || expr2)
int main(){
int i = 1, j=1, k=1;
printf("%d\n", ++i || ++j);
printf("%d %d %d \n", i, j, k);
return 0;
}
Output:
1
2 1 1
Finally, in case of ( expr1 && expr2 ) expr2 only gets executed when expr1 is true or 1, in this case the expr1 is not true hence ++k won't increament. Also following the same above rule for first printf.
int main(){
int i = 1, j=1, k=1;
printf("%d\n", ++i || ++j && ++k);
printf("%d %d %d \n", i, j, k);
return 0;
}
Output:
1
2 1 1
Please see following references as well:
