My recursive function needs to return the smallest vector, only it is always returning 0. What is wrong with my logic?
My role:
int menor(int vet[], int i)
{
int m=0;
if(vet[i] == 0)
{
m=vet[i];
return m;
}
if( vet[i] < m)
{
m=vet[i];
}
return menor(vet,i-1);
}
int main()
{
int vetor[]= {1,2,3,5,7,11,13,17};
m=menor(vetor,7);
printf("%d\n",m );
return 0;
}
You can implement a function that receives a vector of integers and their respective size and returns the smallest value found within this vector, see:
int menor( int vec[], int tam )
{
if( tam == 1 )
return vec[0];
int m = menor( vec + 1, tam - 1 );
return ( vec[0] < m ) ? vec[0] : m;
}
Testing:
#include <stdio.h>
int menor( int vec[], int tam )
{
if( tam == 1 )
return vec[0];
int m = menor( vec + 1, tam - 1 );
return ( vec[0] < m ) ? vec[0] : m;
}
int main( void )
{
int vetor[]= {8,3,5,7,11,13,17,4,8,10};
int tam = sizeof(vetor) / sizeof(int);
int min = menor( vetor, tam );
printf( "%d\n", min );
return 0;
}
Output:
3
See working at Ideone.com
What is happening is that you are zeroing the value of m with each iteration, so it will always return m=0 . An alternative is to make m a function parameter, passed at each iteration. It would look like this:
#include <stdio.h>
int menor(int vet[], int i, int m) {
// Se tiver chegado já ao fim do array para o código
if(i < 0) {
return m;
}
// Verifica se o valor atual é menor que o MENOR valor salvo até então
if(vet[i] < m) {
m = vet[i];
}
// Chama a recursão
return menor(vet, i-1, m);
}
int main() {
int vetor[]= {8,2,3,5,7,11,13,17};
// Adicionamos um parâmetro na chamada, no caso o último valor do vetor.
int m = menor(vetor, 7, vetor[7]);
printf("%d\n",m);
return 0;
}
Notice that we start with a kick, which belongs to the vector, of the smallest value instead of starting with a predefined value. Could the way you did it lead to unexpected behaviors, for example, and if no value were less than 0? 0 would be returned and could not even belong to the searched vector.