Are std::string with null-character possible?

Question

I initialized a C++ string with a string literal and replaced a char with NULL.

When printed with cout << the full string is printed and the NULL char prints as blank.

When printed as c_str the string print stop at the NULL char as expected.

I'm a little confused. Does the action came from cout? or string?

int main(){
  std::string a("ab0cd");
  a[2] = '\0'; // '\0' is null char

  std::cout << a << std::endl; // abcd
  std::cout << a.c_str() << std::endl; // ab
}

Test it online.

I'm not sure whether the environment is related, anyway, I work with VSCode in Windows 10

Replace c with empty character: ''. You are assigning a null character instead — Asesh
– Asesh, Commented Dec 30, 2019 at 9:17
@Asesh There is no "empty character" in C++. Character is a number, there is no concept of "empty number". Empty string there is, of course. — hyde
– hyde, Commented Dec 30, 2019 at 9:28
Note that NULL is not a character, it's an obsolete (legacy C) macro for nullptr. Don't use it for chars. Use '\0' or just plain 0 instead. You are confusing NULL with ASCII character name NUL for this 0 char (which isn't defined as a standard symbol in C++), maybe. — hyde
– hyde, Commented Dec 30, 2019 at 10:00

Lukas-T · Accepted Answer · 2019-12-30 12:03:21Z

5

First you can narrow down your program to the following:

#include <iostream>
#include <string>

int main(){
    std::string a("ab0cd");
    a[2] = '\0'; // replace '0' with '\0' (same result as NULL, just cleaner)

    std::cout << a << "->" << a.c_str();
}

This prints

abcd->ab

That's because the length of a std::string is known. So it will print all of it's characters and not stop when encountering the null-character. The null-character '\0' (which is equivalent to the value of NULL [both have a value of 0, with different types]), is not printable, so you see only 4 characters. (But this depends on the terminal you use, some might print a placeholder instead)

A const char* represents (usually) a null-terminated string. So when printing a const char* it's length is not known and characters are printed until a null-character is encountered.

edited Dec 30, 2019 at 12:03

answered Dec 30, 2019 at 9:24

Lukas-T

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1 Comment

KamilCuk Over a year ago

'\0' (which is equivalent to NULL and 0 - is equivalent to the value. The '\0' has the type char, NULL has the type void* and 0 has the type int.

Orace · Accepted Answer · 2019-12-30 10:03:08Z

2

Contrary to what you seem to think, C++ string are not null terminated.

The difference in behavior came from the << operator overloads.

This code:

cout << a.c_str(); // a.c_str() is char*

As explained here, use the << overloads that came with cout, it print a char array C style and stop at the first null char. (the char array should be null terminated).

This code:

cout << a; // a is string

As explained here, use the << overloads that came with string, it print a string object that internally known is length and accept null char.

edited Dec 30, 2019 at 10:03

answered Dec 30, 2019 at 9:29

Orace

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Comments

user12450543 · Accepted Answer · 2019-12-30 11:22:57Z

0

string end limit (boundary) is not 0 (NULL) like simple char* but its size keep internally in its member data as it's actually user-defined type (an instantiated object) as opposed to primitive type, so

int main(){
  string a("abc0d");
  a[3] = 0; // '\0' is null char
    a.resize(2);
  std::cout << a << std::endl; // ab
  std::cout << a.c_str() << std::endl; // ab
}

i'm sorry change your code to be more comfortable, watch as it results in

ab
ab

good learning: http://www.cplusplus.com/reference/string/string/find/index.html

answered Dec 30, 2019 at 11:22

user12450543

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