Why are my if statements not working consistently? - c++

Question

I'm making a coin toss program for my c++ class and we are required to make a function that flips a coin and prints out if it is heads or tails, and print 10 per line. When I ran the program though the if statements I used to detect if the coin was heads or tails weren't enough to pick from the two.
#include <iostream>
#include <ctime>
using namespace std;
void coinToss(int times);
int main()
{
srand(time(0));
int times;
cout << "How many times would you like to toss the coin?" << endl;
cin >> times;
coinToss(times);
return 0;
}
void coinToss(int times)
{
int toss = 0, count = 0;
for(int i = 0; i < times;i++)
{
toss = rand()%2;
if(toss == 1)//Detects if coin is heads.
{
cout << "H";
}
if(toss == 0)//Detects if coin is tails.
{
cout << "T";
}
else //I had to include this for the program to run, further explanation below the code.
{
cout << "Ya done goofed.";
}
count++; //Counts to ten
if(count == 10) //Skips to the next line if the coin has been tossed ten times.
{
cout << endl;
count = 0;
}
}
}
At one point I replaced the heads or tails with "cout << toss;" and the only numbers returned were 1 and 0. I don't understand how if I'm getting only the two numbers I'm checking for some of them aren't being caught by my if statements.
To complete the assignment I've changed the second if statement into an else statement and everything seems peachy, but I'd really like to understand what's going on here.

Answer 1

What happens with your code is:
Is the result 1 ? Then print H. Keep going. Is the result 0 ? Then print T. Else, if it's not 0, print "Ya done goofed.".
You need to keep your if statements linked together:
if (toss == 1) {
cout << "H";
} else if (toss == 0) {
cout << "T";
} else {
cout << "Ya done goofed.";
}
You won't fall in the else case anymore and will be able to remove it.
As a sidenote, regarding your overall program structure: your coinToss function shouldn't do everything. Your code should be more splitted: a function which returns H or T, a function which calls this function X times as requested by the user and formatting the output would be a good start.
Another small note: your count variable, allowing you to add a new line every 10 flips, could be removed. i % 10 will give you the same result: every ten increments, i % 10 would be equal to 0.

Answer 2

You're probably printing the output properly, then terminating without writing a newline on the last line, and your shell prompts clearing back to the left margin and overwriting your output (clearing the rest of the line to boot). If you have less than 10 tosses, your only line of output may appear lost, otherwise it'll be the last line.
Try adding an extra std::cout << '\n'; before main returns.
(Separately, you can say std::cout << "HT"[rand() % 2];, or std::cout << (rand() % 2 ? 'H' : 'T'); and do away with the ifs, but it's no big deal... whatever's clearest for you at this stage)

Answer 3

Well, rand()%2 will produce only two numbers: 1 and 0, this seems to be in line with your task as a coin is a boolean number generator, isn't it? :)
Therefore this seems to do the job you are looking for:
#include <iostream>
#include <ctime>
using namespace std;
void coinToss(int times);
int main()
{
srand(time(0));
int times;
cout << "How many times would you like to toss the coin?" << endl;
cin >> times;
coinToss(times);
return 0;
}
void coinToss(int times)
{
int toss = 0, Count = 0;
for(int i = 0; i < times;i++)
{
toss = rand() % 2;
// Choose:
cout << ((toss) ? "H" : "T"); // if you want a character
// or
cout << toss; // if you want the number
Count++; //Counts to ten
if(Count == 10) //Skips to the next line if the coin has been tossed ten times.
{
cout << endl;
Count = 0;
}
}
}

Answer 4

if(toss == 1)//Detects if coin is heads.
{
cout << "H";
}
else if(toss == 0)//Detects if coin is tails.
{
cout << "T";
}
You need to use else-if statement. You also need not use else after the toss==0 because rand()%2 will either be 0 or 1. There is no third option.

Answer 5

rand() returns a pseudo-random integral number in the range between 0 and RAND_MAX. And, rand() % 2 will be 0 or 1. So, there would be:
if(toss == 1)//Detects if head
{
cout << "H";
}
else // tail
{
cout << "T";
}

Answer 6

I don't think there is anything wrong with this. Well not that I can see... If I add some debug then I see what I think you're expecting...
#include <iostream>
#include <ctime>
using namespace std;
void coinToss(int times);
int main() {
srand(time(0));
int times;
cout << "How many times would you like to toss the coin?" << endl;
cin >> times;
coinToss(times);
return 0;
}
void coinToss(int times) {
int toss = 0, count = 0;
for(int i = 0; i < times;i++) {
toss = rand() % 2;
cout << "Toss: " << toss << endl;
if(toss == 1)//Detects if coin is heads.
{
cout << "H (" << toss << ")" << endl;
}
if(toss == 0)//Detects if coin is tails.
{
cout << "T (" << toss << ")" << endl;
}
count++; //Counts to ten
if(count == 10) //Skips to the next line if the coin has been tossed ten times.
{
//cout << endl; count = 0;
}
}
}
And compile it
g++ coin_toss.cc
And run it
./a.out
How many times would you like to toss the coin?
4
Toss: 1
H (1)
Toss: 0
T (0)
Toss: 0
T (0)
Toss: 0
T (0)
Then this is exactly what I expect or am I missing something?
You don't need an "if else if" statement.

Answer 7

You can also use a switch:
switch( rand() % 2 )
{
case 0:
cout << "T";
break;
case 1:
cout << "H";
break;
default:
cout << "oops you goofed!;
}
// continue within for loop
If you "forgot" the break after case 1 you would again get the "oops you goofed!" message after each head toss.