null character inside string

Question

From Rules for C++ string literals escape character ,Eli's answer

std::string ("0\0" "0", 3) // String concatenation

works because this version of the constructor takes a char array; if you try to just pass "0\0" "0" as a const char*, it will treat it as a C string and only copy everything up until the null character.

Does that mean space isn't alloted for entire string , ie the string after \0 is written on unalloted space ?

Moreover the above question is for c++ string, I observed same behaviour for c strings too . Are c and c++ strings same when I add null char in middle of string during declaration ?

What are you trying to do here: Why do you need to use strings instead of just character arrays? — hugomg
– hugomg, Commented Jul 28, 2013 at 8:34
There is a set of strings which could be represented as a array of pointers to char[] , but is represented as char[] with \0 separating strings and int[] to store offsets of strings — Zxcv Mnb
– Zxcv Mnb, Commented Jul 28, 2013 at 8:41

Oswald · Accepted Answer · 2013-07-28 08:41:26Z

2

The char array is copied into the new object. If you don't specify, how long the char array is, C++ will copy until the first null character. How much additional space is allocated is outside the scope of the specification. Like vectors, strings have a capacity that can exceed the amount required to store the string and allows to append characters without relocating the string.

edited Jul 28, 2013 at 8:41

answered Jul 28, 2013 at 8:34

Oswald

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7 Comments

Zxcv Mnb Over a year ago

so something like char w[]="qwe\0""0uqwe"; printf("%s\n%s",w,w+4); is wrong ?

Oswald Over a year ago

char w[]="qwe\0""0uqwe" casts a const char* to char[]. That's wrong.

Casey Over a year ago

The string capacity isn't completely outside the scope of the spec: you're guaranteed that capacity >= size.

Oswald Over a year ago

"How much additional space is ...". Better?

Zxcv Mnb Over a year ago

So is size here size of "qwe" or size of "qwe\0""0uqwe" ?

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P0W · Accepted Answer · 2013-07-28 08:55:51Z

std::string constructor that takes a const char* assumes the input is a C string. C strings are '\0' terminated and thus stops when it reaches the '\0' character.

If you really want to play you need to use the constructor that builds the string from a char array (not a C-String).

STL sequence containers exceed the amount required to store automatically

Example :

int main () { string s="Elephant"; s[4] ='\0'; //C++ std::string is NOT '\0' terminated cout<<s<<endl; //Elep ant string x("xy\0ab"); // C-String assumed. cout<<x; //xy return 0; }

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