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From Rules for C++ string literals escape characterRules 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 ?

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 ?

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 ?

this is purely C++
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null character inside string

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 ?