Why does the "for" loop not work and program stops?

Question

When I compile the program it seems to work normally but, when it asks for the "cuadrados", the program just stops:

int main(){

int *ptrs, suma=0, n, i, m;

printf("cuantos numeros al cuadrado quiere?: ");
scanf("%i", &n);

int numero[n], cubo[n];
*ptrs=numero

Here is when the program stops:

for(i=0;i<n;i++){
printf("escriba los cuadrados: "); 
scanf("%i", numero[i]);

}


printf("Cuantos numero al cubo quiere?: ");
scanf("%i", m);

 for(i=0;i<n;i++){
printf("escriba los cubos: ");
scanf("%i", cubo[i]);
}

I had this issue before but I can't understand why it does not keep running; it just asks for 1 number and then stops with no error or warning.

Answer 1

for(i=0;i<n;i++){
printf("escriba los cuadrados: "); 
scanf("%i", &numero[i]);
}

printf("Cuantos numero al cubo quiere?: ");
scanf("%i", &m);

 for(i=0;i<n;i++){
printf("escriba los cubos: ");
scanf("%i", &cubo[i]);
}

You need to give address of the variable in scanf using & operator

Answer 2

This will crash because scanf needs a pointer-to-int , not an int :

scanf("%i", numero[i]);

This will not:

if (scanf("%i", &numero[i]) == 1) {
    /* do something */
}
else {
     /* scanf failed */
}

Note that not testing the scanf return value is always asking for trouble. As is not using your compiler's maximum warning level and turning all warnings into errors.