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bearchen
2 New Member
Hello
I'm dividing consecutive prime numbers (larger over lesser) and I get 1.0000(....) for all of them.
Does it have anything to do with runnig gcc on windows? I've tried the calculator applet that comes with the OS and it works properly...
The program prints a column with the order, a second with each prime, then the difference between two consecutives, the cumulative sum and the last one was supposed to be the ratio between two consecutive primes.
Here's part of the code:
(main)
(........)
for ( m = 0; m <= MAX_DIM; m++ ){
if ( vect_primos[ m ] != 0 && vect_primos[ m ] < 10000 )
printf ( "%6d%8d%12d%12d %17.6lf\n", ( m + 1 ), vect_primos[ m ], ( vect_primos[ m ] - vect_primos[ m - 1 ] ), cum_sum( &vect_primos[ m ] ), divisao( &vect_primos[ m ] ) );
}
return 0;
}
------------------------------------------
int cum_sum( int v[ ]){
int i;
for ( i = 0; i <= MAX_DIM; i++ ){
acumulado[ i ] = acumulado[ i ] + v[ i ];
return acumulado[ i ];
}
return 0;
}
----------------------------------------------------------
double divisao( int vec[ ] ){
int i;
for ( i = 0; i <= MAX_DIM; i++ ){
div[ i ] = vec[ i + 1 ] / vec[ i ];
return div[ i ];
}
return 0;
}
Any suggestions?
Many thanks in advance.
Sep 5 '09 #1
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2 3283 
donbock 2,426 Recognized Expert Top Contributor
@bearchen
For example I'll use consecutive primes 5 and 3, the value of 5/3 is different than 5.0/3.0. The first expression uses integer division; the second uses floating-point arithmetic.
By the way, function divisao references the passed array element and the next one. Any time you infer the address of a parameter like this you need to be extra careful that your loop-termination criteria stops at the right time. You don't want to pass divisao the address of the last item in the array.
Sep 5 '09 #2
reply
bearchen 2 New Member
Thanks, I keep forgetting these basic principles. I promoted the array vec[ ] to float and now it works.
After some modifications and patches, here's the end result:
Sep 5 '09 #3
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