Removing EOL from a C++ string but preserving the '\0'

Question

This is probably a dumb question but if I have a C++ string as follows,

str = "the cat is black\n"

If I remove the last character, namely '\n', does this leave me with a null terminated string? Does the statement below work in creating a null terminated char * from the original string str?

char *c = str.erase(str.size()-1).c_str();

You remove the last character by moving a `'\0'` there, because that `'\0'` terminates the string. — , Feb 18 '16 at 18:09
@Rhymoid: That's a bit of a simplification... the string will also typically have a length field which is updated, and you can have an `\0` in the middle of your string, if you want. — Dietrich Epp, Feb 18 '16 at 18:11
For C `'\n'` removal, see http://stackoverflow.com/q/2693776/2410359 — chux - Reinstate Monica, Feb 18 '16 at 18:29
`std::string::erase` keeps the string internals all sorted correctly (as do all member functions, etc.), so `.c_str()` will behave as with any `std::string` object (i.e. null terminated). — crashmstr, Feb 18 '16 at 18:44

score 5 · Answer 1 · answered Feb 18 '16 at 18:07

5

Yes, std::string::c_str() will always return a null-terminated string.

http://www.cplusplus.com/reference/string/string/c_str/

http://en.cppreference.com/w/cpp/string/basic_string/c_str

answered Feb 18 '16 at 18:07

Dan Korn

1,274
9
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1

Thanks!! I'm understanding that as long as it's a c.str() then the '\0' termination is automatic, – MikeRin Feb 21 '16 at 23:34

Vlad from Moscow · Answer 2 · 2016-02-18T18:21:14.517

If it is a C string as for example

char str[] = "the cat is black\n";

you can do this like

str[strcspn( str, "\n" )] = '\0';

If it is an object of type std::string as for example

std::string str( "the cat is black\n" );

you can write

if ( str.back() == '\n' ) str.erase( str.size() - 1 );

or

if ( !str.empty() && str.back() == '\n' ) str.erase( str.size() - 1 );

This statement

const char *c = str.erase( str.size() - 1 ).c_str();
^^^^^

is also valid provided that the string contains the new line character or empty.

Removing EOL from a C++ string but preserving the '\0'

2 Answers2