Reputation: 311
I am fairly new to C++ and coding in general. I am trying to make just a basic little multiple choice type game for practice but I have run into a conundrum.
The program isn't outputting what I want it too. Here is the code:
#include <iostream>
#include <cstdlib>
#include <ctime>
#include <string>
using namespace std;
void sword(int damage);
void fists(int damage);
static int enemyHealth = 250;
int main() {
srand(time(0));
string SOF; //Abreveation for "Sword Or Fists"
cout << "You will be fighting a mean bad guy. Are you using a sword, or your fists?\n";
while (SOF != "sword" && SOF != "fists"){
cout << "Please enter your choice of either 'sword' or 'fists': ";
cin >> SOF;
}
cout << "Okay! Time to fight! \n";
if (SOF == "fists") {
void fists();
}
else if (SOF == "sword") {
void sword();
}
else{ (NULL); }
cout << "Congratulations! You have vanquished that foul beast!\n";
system("pause");
}
//This is for when the user chooses 'sword'
void sword(int damage = rand() % 100 + 50) {
while (enemyHealth > 0){
cout << "You deal " << damage << " damage with your sharp sword. \n";
enemyHealth -= damage;
}
}
//This is for when the user chooses 'fists'
void fists(int damage = rand() % 10 + 4) {
while (enemyHealth > 0){
cout << "You deal " << damage << " damage with your womanly fists. \n";
enemyHealth -= damage;
}
}
The first part works fine, but when I enter my choice of either "fists"
or "sword"
the output is:
Okay! Time to fight!
Congratulations! You have vanquished that foul beast!
But I want it to output the damage being done with either fists or sword.
If I could get some help with that, it would be amazing. Thanks!
Upvotes: 1
Views: 68
Reputation: 18474
void fists();
is a declaration, not a call, change to fists();
and sword();
Other things to look at:
main
(or just move whole functions there)SOF
looks loke it is a #define
d constant or such.Upvotes: 3
Reputation: 18974
To call the function, don't write void fists();
, just
fists();
(What you have is a declaration, which has no useful effect here, rather than a call.)
Upvotes: 1