CODEWITHSUNDEEP
c++c++11
Aryan Agarwal
Aryan Agarwal

Reputation: 23

Garbage when converting character to string using concatenation

I am converting a character to a string by concatenating it with an empty string (""). But it results in undefined behaviour or garbage characters in the resultant string. Why so?

char c = 'a';
string s = ""+c;
cout<<s<<" "<<s.size()<<"\n";

Upvotes: 0

Views: 355

Answers (2)

Peter Lawrence
Peter Lawrence

Reputation: 54

An interesting issue, which looks like an issue which the compiler does not detect at compilation.
I have tried this with GNU C++ 10.2 on Cygwin and your code snippet generates this.

0200312 (Fedora Cygwin 9.3.0-1) 31

if you replace char c = 'c'; with char c = '\0'; the output is

 0

So looks like some kind of memory violation issue.

As you probably know, the standard method of appending a char is to use push_back

string s = "";
s.push_back(c);
cout << s << " " << s.size() << "\n";

Alternatively you can use the operator +=

string s = "";
s += c;
cout << s << " " << s.size() << "\n";

Upvotes: 0

Thomas Matthews
Thomas Matthews

Reputation: 57728

Let's look at your snippet, one statement or a line at a time.

char c = 'a';

This is valid, assigning a character literal to a variable of type char.
Note: since c is not changed after this statement, you may want to declare the definition as const.

string s = ""+c;

Let's refactor:
std::string s = ("" + c);

Let's add type casting to make the point more clear:
std::string s = ((const char *) "" + (char) c);

The order of operations is resolve all expressions on the right hand side (rhs) of the assignment operator before assigning to the string variable.

There is no operator+() in C++ that takes a const char * and a single character.
The compiler is looking for this function: operator+(const char *, char).
This is the primary reason for the compilation errors.

cout<<s<<" "<<s.size()<<"\n";
The string assignment and creation failed, thus s is empty and s.size() is zero.

Upvotes: 1

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