Answered step by step
Verified Expert Solution
Question
1 Approved Answer
INPUT OUTPUT John Doe 333224444 DoeJ 123Password John Doe xxxxxxxxx DoeJ xxxxxxxxxxx Jane Smith 296487534 SmithJ 123456789Hey Jane Smith xxxxxxxxx SmithJ xxxxxxxxxxxx Write a program
INPUT | OUTPUT |
John Doe 333224444 DoeJ 123Password | John Doe xxxxxxxxx DoeJ xxxxxxxxxxx |
Jane Smith 296487534 SmithJ 123456789Hey | Jane Smith xxxxxxxxx SmithJ xxxxxxxxxxxx |
Write a program that reads in a line consisting of a student's name, Social Security number, user ID, and password. The program outputs the string in which all the digits of the Social Security number and all the characters in the password are replaced by x. (The Social Security number is in the form 000-00-0000, and the user ID and the password do not contain any spaces.) Your program should not use the operator [] to access a string element. Use the appropriate functions below.
Expression | Effect |
strVar.at(index) | Returns the element at the position specified by index |
strVar[index] | Returns the element at the position specified by index |
strVar.append(n, ch) | Appends n copies of ch to strVar, where ch is a char variable or a char constant |
strVar.append(str) | Appends str to strVar |
strVar.clear() | Deletes all the characters in strVar |
strVar.compare(str) | Returns 1 if strVar > str returns 0 if strVar == str; returns 1 if strVar < str |
strVar.empty() | Returns true if strVar is empty; otherwise it returns false |
strVar.erase() | Deletes all the characters in strVar |
strVar.erase(pos, n) | Deletes n characters from strVar starting at position pos |
strVar.find(str) | Returns the index of the first occurrence of str in strVar. If str is not found, the special value string::npos is returned |
strVar.find(str, pos) | Returns the index of the first occurrence at or after pos where str is found in strVar |
strVar.find_first_not_of(str, pos) | Returns the index of the first occurrence of any character of strVar in str. The search starts at pos |
strVar.insert(pos, n, ch) | Inserts n occurrences of the character ch at index pos into strVar; pos and n are of type string::size_type; and ch is a character |
strVar.insert(pos, str) | Inserts all the characters of str at index pos into strVar |
strVar.length() | Returns a value of type string::size_type giving the number of characters in strVar |
strVar.replace(pos, n, str) | Starting at index pos, replaces the next n characters of strVar with all the characters of str. If n > length of strVar, then all the characters until the end of strVar are replaced |
strVar.substr(pos, len) | Returns a string which is a substring of strVar starting at pos. The length of the substring is at most len characters. If len is too large, it means "to the end" of the string in *strVar |
strVar.size() | Returns a value of type string::size_type giving the number of characters in strVar |
strVar.swap(str1) | Swaps the contents of strVar and str1. str1 is a string variable |
Step by Step Solution
There are 3 Steps involved in it
Step: 1
Get Instant Access to Expert-Tailored Solutions
See step-by-step solutions with expert insights and AI powered tools for academic success
Step: 2
Step: 3
Ace Your Homework with AI
Get the answers you need in no time with our AI-driven, step-by-step assistance
Get Started