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You can use cin but the cin object will skip any leading white space (spaces, tabs, line breaks), then start reading when it comes to the first non-whitespace character and then stop reading when it comes to the next white space. In other words, it only reads in one word at a time.
Whitespace in C++cout<<"Hello"; cout << "Hello"; cout << "Hello" ; cout << "Hello"; The exceptions where C++ compiler takes whitespace in consideration is inside quotes and for operator detection. So whenever you put in a string, c++ takes note of the whitespace.
Once the character is equal to New-line ('\n'), ^ (XOR Operator ) gives false to read the string. So we use “%[^\n]s” instead of “%s”. So to get a line of input with space we can go with scanf(“%[^\n]s”,str);
It doesn't "fail"; it just stops reading. It sees a lexical token as a "string".
Use std::getline:
#include <string>
#include <iostream>
int main()
{
std::string name, title;
std::cout << "Enter your name: ";
std::getline(std::cin, name);
std::cout << "Enter your favourite movie: ";
std::getline(std::cin, title);
std::cout << name << "'s favourite movie is " << title;
}
Note that this is not the same as std::istream::getline, which works with C-style char buffers rather than std::strings.
Update
Your edited question bears little resemblance to the original.
You were trying to getline into an int, not a string or character buffer. The formatting operations of streams only work with operator<< and operator>>. Either use one of them (and tweak accordingly for multi-word input), or use getline and lexically convert to int after-the-fact.
You have to use cin.getline():
char input[100];
cin.getline(input,sizeof(input));
The Standard Library provides an input function called ws, which consumes whitespace from an input stream. You can use it like this:
std::string s;
std::getline(std::cin >> std::ws, s);
Use :
getline(cin, input);
the function can be found in
#include <string>
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