This question already has answers here:
What do single quotes do in C++ when used on multiple characters?
(5 answers)
Closed 6 years ago.
I have displayed a statement in C++ program with single quote and answer which i got was the random numbers where as when i used the double quotes in C++ it displayed me the statement.
cout << 'Hello world'; //gave me the random numbers
cout << "Hello world"; //displayed me the statement i.e Hello world
Why This happened pls do let me know and what those random numbers were at the time of execution ?
Single quotes are used for characters:
char c = 'A';
In case you want to print a string which is more than one character, you use double quotes:
cout << "Hello World";
You are trying to print an array of characters, which require double quotes.
char str[3] = "abc"; //an array of characters, use double quotes
str[0] = 'x'; //set a character in this array, use single quote
Related
This question already has answers here:
C++ character concatenation with std::string behavior. Please explain this
(3 answers)
Closed 1 year ago.
cout<<"#" + 'a'<<endl;
string s = "#";
s += 'a';
cout<<s<<endl;
I am not able to figure out how the typecasting is working in the case "#" + 'a'
In cpp string works like an array of characters, so when you assign s = '#' it compiles like this:
s[0] = '#'
and in the second line it actually compiles like this:
s[1] = 'a'
finally, s is:
#a
This question already has an answer here:
C++ string and string literal comparison
(1 answer)
Closed 1 year ago.
Question - The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word s from Berlandish into Birlandish as t. Help him: find out if he translated the word correctly.
Input -
The first line contains word s, the second line contains word t. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output -
If the word t is a word s, written reversely, print YES, otherwise print NO.
When I write this code, the output is wrong -
int main(){
char s[100000],a[100000];
cin >> s >> a;
strrev(s);
if(s==a){
cout << "YES";
}else{cout << "NO";}
}
But when I write this code, the output is correct -
int main(){
char s[100000];
string a;
cin >> s >> a;
strrev(s);
if(s==a){
cout << "YES";
}else{cout << "NO";}
}
Why is it like this, is there a rule that a character array cannot be compared to another character array and if so, how can it be compared to a string?
Remember that arrays naturally decay to pointers to their first elements, and it's such pointers that you are comparing.
In short, what you're really doing is:
if(&s[0] == &a[0])
And those two pointers will never be equal.
To compare the contents of character arrays, you need to use strcmp() or similar function instead, eg:
if(strcmp(s, a) == 0)
Since you're programming in C++, please use std::string for all your strings. There are overloads for the == operator that do the right thing if you have std::string values.
This question already has answers here:
How can I repeat a string a variable number of times in C++?
(10 answers)
Closed 2 years ago.
one is 2, and ans is "000000".
string ans = "000000";
ans += string("1", one);
cout<<ans<<endl;
The output is:
0000001�
But I want the output:
00000011
What am I doing wrong?
string("1", one) does not do what you think it does. It does not duplicate the "1" string one number of times. It instead copies the 1st one number of chars from "1", which in this case is the '1' character and the '\0' null-terminator that follows it, which is where the � is coming from in the output. That is not what you want.
Use string(one, '1') instead. That will duplicate the '1' character one number of times, like you want, eg:
ans = "000000";
ans += string(one, '1');
cout << ans << endl;
Just use c++ strings and use + operator to catenate strings.
This question already has answers here:
Change string by index
(3 answers)
Closed 7 years ago.
string l;
cin>>l[0];
cout<<l;
input:a
output:
According to me the code must print the value of l[0] but why is there no output?
You need to do
cin>>l;
In your current state of code, when you try to access l[0] you are trying to access a memory location which may or may not be there. cin >> l[0] doesn't changes the size of the string which remains 0.
So when you try to do cout << l you are effectively printing an empty string.
Another alternative is
string s;
s.resize(1);
You have an empty string. There are no characters in it, including 0th character, you trying to read into. You need to actually add characters in string:
std::string l;
l.push_back('\0'); //Or any other character
std::cin >> l[0];
This question already has answers here:
How do you construct a std::string with an embedded null?
(11 answers)
Closed 8 years ago.
I'm trying to read a large buffer from a socket which uses \0 to delimit pieces of data and \n to delimit lines.
I thought getline() would be an easy way to get each line but it's behaving strangely.
I'm using \n as the delimiter in getline().
string line;
string test1 = "aaa,123\nbbb\nccc,456\n";
stringstream ss1(test1);
while(std::getline(ss1, line, '\n')) {
cout << line << endl;
}
// outputs:
// aaa,123
// bbb
// ccc,456
string test2 = "aaa\0123\0\nbbb\0\nccc\0456\0\n";
stringstream ss2(test2);
while(std::getline(ss2, line, '\n')) {
cout << line << endl;
}
// outputs:
// aaa
// 3
Why is this happening in test2? Where is the 3 coming from? Must I remove the \0 to make this work? Is there an easier/better way to mark strings in my buffer when I do a socket recv()?
\0 in a special symbol. It shows when the string ends.
For example, if you type in "a string", the compiler automatically adds a \0 on the end, which signifies the end of the string. However, it is legal to have a \0 in the middle of the string, it just means that everything after it is ignored.
So basically, any operation you do on the string, not just the getline, will treat the string as "aaa", ignoring everything after the first \0 that is found. But...
As #Fred Larson points out
Oh, I see where the 3 comes from. The first \0 isn't a null, it's the start of \012, which is a carriage return. Then the 3 follows.
So actually, the string is being treated as "aaa\n3". Which is why you get the output you do.
Edit: And thanks to Galik, I will also add that these rules I mention might only apply to a string literal / c-string. It may be a different case with std::strings, in which the length of the string is known ahead of time.
\0 is the standard string terminator symbol. As such, you may either read character by character or avoid \0 as delemeters