+ 3
is there a way I can store all the values of x in this short code without having to display them using cout? int main() { for (int x=0; x<5; x++){ //how do i store the all the values of x without having to display them?// }
5 Réponses
+ 7
You may use an array to do so.
int main ()
{
int array[5];
for (int x = 0; x < 5; x++)
{
array[x] = x;
}
return 0;
}
... Although I fail to see the point of doing so.
Now, the set of integers stored in array would be as follows:
{0, 1, 2, 3, 4}
+ 6
Here goes. I am using switch statements though, to make it shorter.
int x = 0;
int choice1 = 0;
int weaponpoint;
int weap_sum;
int array[5];
srand (time (0));
for (x = 0; x < 5; x++)
{
choice1 = 1+ rand() * 10;
switch (choice 1)
{
case 1:
weaponpoint = 10;
array[x] = weaponpoint; break;
case 2:
weaponpoint = 20;
array[x] = weaponpoint; break;
// similar for the rest of the cases
case 10:
weaponpoint = 100;
array[x] = weaponpoint; break;
}
}
for (int j = 0; j < 5; j++)
{
weap_sum += array[j];
}
cout << "weap_sum = " << weap_sum << endl;
cout << "weap_sum = ";
for (int k = 0; k < 4; k++)
{
cout << array[k] << " + ";
}
cout << array[4];
+ 3
thanks guys for your answers, but to be more specific, this is the code.
I want to store each value of weaponpoint . And then after the looping, I want to display the addition of the weapon point values in the form (eg 50+20+60+10+30). how will I be able to store and then obtain each weaponpoint value?
srand(time(0));
for(int x= 0;x<5;x++){
choice1 =1+(rand()%10) ;
if (choice1 ==1){
weaponpoint =10;
}
else if (choice1 ==2){
weaponpoint=20;
}
else if (choice1 ==3){
weaponpoint=30;
}
else if (choice1 ==4){
weaponpoint=40;
}
else if (choice1 ==5){
weaponpoint =50;
}
else if (choice1 ==6){
weaponpoint=60;
}
else if (choice1 ==7){
weaponpoint=70;
}
else if (choice1 ==8){
weaponpoint=80;
}
else if (choice1 ==9){
weaponpoint=90;
}
else if (choice1 ==10){
weaponpoint =100;
}
0
store the values in an array
0
cout is used to DISPLAY values not to store them arrays can simply store them otherwise if you explained why or what you want to do I would be able to help you