Consider the function, where the programmer inadvertently left out the ampersand in the definition of parRef. What is the output of the code?

Question

```
void doStuff(int parValue, int parRef)
{
parValue = 100;
cout << “parValue in call to doStuff = “
<< parValue << endl;
parRef =222;
cout << “parRef in call to doStuff = “
<< parRef << endl;
}
```

and consider the call, which we assume is in a complete and correct program
```
int n1 = 1, n2 =2;
doStuff(n1, n2);
cout << “n1 after call to doStuff = “ << n1 << endl;
cout << “n2 after call to doStuff = “ << n2 << endl;
```

a. parValue in the call to doStuff = 100;
parValue in the call to doStuff = 222;
n1 after function call = 1;
n2 after function call = 222
b. parValue in the call to doStuff = 100;
parValue in the call to doStuff = 222;
n1 after function call = 1;
n2 after function call = 2
c. x parValue in the call to doStuff = 100;
parValue in the call to doStuff = 222;
n1 after function call = 100
n2 after function call = 2
d. parValue in the call to doStuff = 100;
parValue in the call to doStuff = 222;
n1 after function call = 100
n2 after function call = 222

Answer 1

b. parValue in the call to doStuff = 100;
parValue in the call to doStuff = 222;
n1 after function call = 1;
n2 after function call = 2

The omission of the ampersand causes the function’s only action to be the output generated by its output statements. All the rest suggest that a function with call-by-value parameters can cause changes in the argument. This cannot happen.