ignoring return value of ‘int scanf(const char*, ...)’, declared with attribute warn_unused_result [-Wunused-result]? - g++

Question

When I compiled the following program like:
g++ -O2 -s -static 2.cpp it gave me the warning ignoring return value of ‘int scanf(const char*, ...)’, declared with attribute warn_unused_result [-Wunused-result].
But when I remove -02 from copiling statement no warning is shown.
My 2.cpp program:
#include<stdio.h>
int main()
{
int a,b;
scanf("%d%d",&a,&b);
printf("%d\n",a+b);
return 0;
}
What is the meaning of this warning and what is the meaning of -O2 ??

Answer 1

It means that you do not check the return value of scanf.
It might very well return 1 (only a is set) or 0 (neither a nor b is set).
The reason that it is not shown when compiled without optimization is that the analytics needed to see this is not done unless optimization is enabled. -O2 enables the optimizations - http://gcc.gnu.org/onlinedocs/gcc/Optimize-Options.html.
Simply checking the return value will remove the warning and make the program behave in a predicable way if it does not receive two numbers:
if( scanf( "%d%d", &a, &b ) != 2 )
{
// do something, like..
fprintf( stderr, "Expected at least two numbers as input\n");
exit(1);
}

Answer 2

I took care of the warning by making an if statement that matches the number of arguments:
#include <iostream>
#include <cstdio>
using namespace std;
int main() {
int i;
long l;
long long ll;
char ch;
float f;
double d;
//6 arguments expected
if(scanf("%d %ld %lld %c %f %lf", &i, &l, &ll, &ch, &f, &d) == 6)
{
printf("%d\n", i);
printf("%ld\n", l);
printf("%lld\n", ll);
printf("%c\n", ch);
printf("%f\n", f);
printf("%lf\n", d);
}
return 0;
}