Reputation: 225
I've made a print function in C++ that should be a nicer alternative to std::cout
, and it is very similar to the printf
function in C.
The problem is that it receives a variable number of parameters, but uses them incorrectly.
Here's the source code:
void print(string s, ...)
{
int len = s.length();
int pSum = 0;
for (int i = 0; i < len; i++)
{
if (s[i] == '%' && i < len - 1)
{
char c = s[i + 1];
if (c == 'd' || c == 'i' || c == 's') pSum++;
}
}
va_list ap; // Initialization of the VA list
va_start(ap, pSum); // Start iterating through the VA list
int r;
for (int i = 0; i < len; i++)
{
if (s[i] == '%' && i < len - 1) // Check for % in the string
{
char c = s[i + 1];
if (c == 'i')
{
r = va_arg(ap, int);
cout << r;
i++;
}
}
else cout << s[i];
}
va_end(ap); // Stop iterating over the VA list
fflush(stdout); // Flush standard output stream
}
I tested this code that with that function call:
print("The number %i is the only even prime number.", 2);
And the output was:
The number -858993460 is the only even prime number.
Why does this happen? I have tried debugging but have absolutely no idea where does that problem come from.
Upvotes: 0
Views: 425
Reputation: 218090
The second argument of va_start
should be the last named parameter of the function so use:
va_start(ap, s);
Upvotes: 2