+ 6

Parsing string integer value to normal integer?

How to parse a string integer to a normal integer? eg: string str="ahd14sjd"; int a; for(int I=0;I<str.length();I++) { if(str[I]=='1') a=str[I]; } cout<<a; In above example, The value of 'a' will be ASCII value of 1. but I want to assign 'a' as 1 and print 'a' as 1. how can I do that? (I.e) parsing string integer to a integer? I hope you understood my question.

24th Feb 2017, 7:18 PM
Mr.Robot
Mr.Robot - avatar
3 Answers
+ 3
string str="ahd14sjd"; int a=0; int c=1; for(int i=str.length();i>=0;i--){ if(str[i]>='0'&&str[i]<='9'){ a += (str[i]-'0')*c; c *= 10; } } cout << a;
24th Feb 2017, 8:50 PM
Š®Ń€ŠøŠ¹ Š˜ŃŠ°ŠµŠ²
Š®Ń€ŠøŠ¹ Š˜ŃŠ°ŠµŠ² - avatar
+ 1
The ascii value of 'a' is 97 as an int. If you're trying to get the Integer value of a character, all you need to do is cast the char to an int. int('a')
24th Feb 2017, 8:18 PM
ChaoticDawg
ChaoticDawg - avatar