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cin.getline ()

This is the first part of my code #include <iostream> using namespace std; int main() { int n; cin >> n; cin.ignore (n); char asli [100]; cin.getline (asli,100); cout << asli; Its supposed to get a number n which Im gonna need later and then get a string the reason I used cstyle string is because I want to use cin.getline()، but when I input "zohal" its output is "al" instead of zohal I used different example it never gets the first three characters, why?!

12th Dec 2019, 2:59 PM
Zohal
Zohal - avatar
2 Answers
+ 7
Reason: This👉cin.ignore(n); Assume, input as: 4 zohal Firstly, the computer will assign value 4 to n. As you specified n number of characters to be ignored through cin.ignore(n) predefined function, first 4 characters will be ignored (including spaces/line) So that, _(1 space)zoh(3 characters) will be ignored. Hence, Output will be: al Solution: just replace cin.ignore(n); with any of the following lines below. • cin.ignore(); • cin.ignore(1, '\n'); • cin.ignore(1); Hope this helps. :)
13th Dec 2019, 4:43 PM
Vinay_GB
Vinay_GB - avatar
0
Nobody knows tge reason??
12th Dec 2019, 4:02 PM
Zohal
Zohal - avatar