Reputation: 9614
//sLine is the string
for(int l = 0; l < sLine.length(); l++)
{
string sNumber;
if(sLine[l] == '-')
{
sNumber.push_back(sLine[l]);
sNumber.push_back(sLine[l + 1]);
l++;
}
else if(sLine[l] != '\t')
{
sNumber.push_back(sLine[l]);
}
const char* testing = sNumber.c_str();
int num = atoi(testing);
cout << num;
}
I have this for-loop which checks each character of the string and converts every number in this string to be a int. But for some reason, the atoi function is doing it twice so when I cout it, it displays it twice for some reason... Why is that?
example:
INPUT
3 3 -3 9 5
-8 -2 9 7 1
-7 8 4 4 -8
-9 -9 -1 -4 -8
OUTPUT
3030-309050
-80-20907010
-70804040-80
-90-90-10-40-80
Upvotes: 1
Views: 1689
Reputation: 3065
That seems like a painful way to recreate the wheel. You'd be better off using a stringstream to parse this.
std::stringstream strm(sLine);
int num;
while(strm >> num)
{
std::cout << num << std::endl;
}
Upvotes: 0
Reputation: 66912
It's displaying a zero for all nonrecognized characters, because atoi
returns 0
when given a non-numeric string (like a space!)
However, what you want to do, is shockingly simple:
std::stringstream ss(sLine);
int num;
while(ss >> num) {
cout << num;
}
Upvotes: 10
Reputation: 35584
You output extra 0
for the characters which are not digits. The problem is that atoi
returns 0 when it cannot convert the input, so your whitespaces are printed as zeroes.
Upvotes: 0
Reputation: 42109
Move this:
const char* testing = sNumber.c_str();
int num = atoi(testing);
cout << num;
Below the last }
in the code you pasted, i.e. out of the for-loop. Currently you get a separate printout for every character in sLine
because it's executed on every iteration of the loop. (The last character in sLine
may be a linefeed so this can occur even if you think you wrote only one digit.)
Edit: Also move the declaration of sNumber
above the for-loop.
You may also want to change if (sLine[l] == '-')
to if (sLine[l] == '-' && (l + 1) < sLine.length())
so you don't access beyond the end of the string if the dash is the final character on the line.
You may also want to rename the variable l
to something that looks less like a 1
. =)
You may also want to rethink if this is the right way to do this at all (usually if a simple thing gets this complicated, chances are you're doing it wrong).
Upvotes: 0