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Why is .c_str() not adding '' at the end of a string?



2019 Community Moderator ElectionHow do you construct a std::string with an embedded null?How do I iterate over the words of a string?Why can templates only be implemented in the header file?Why is “using namespace std” considered bad practice?string::c_str queryEasiest way to convert int to string in C++Why are elementwise additions much faster in separate loops than in a combined loop?Why does changing 0.1f to 0 slow down performance by 10x?Why is reading lines from stdin much slower in C++ than Python?Why is it faster to process a sorted array than an unsorted array?Why should I use a pointer rather than the object itself?










3















When running this code I get "Hello Worldaaa", but according to the job of the .c_str() function, the string a should be "Hello World" and string b should be "Hello Worldaaa" so "aaa" should not show up in the output.



#include <bits/stdc++.h>

using namespace std;

int main()

string a = "Hello World";
a = a.c_str();
string b = a + string("aaa");

cout << b;

return 0;










share|improve this question

















  • 3





    "Returns a pointer to a null-terminated character array with data equivalent to those stored in the string." It does, but using it in the std::string operator+(const char*, const std::string&) ignores the terminating '' character to produce the result.

    – πάντα ῥεῖ
    Mar 6 at 17:46






  • 2





    it's not actually .c_str(), it's actually the constructor that's confusing you.

    – Mooing Duck
    Mar 6 at 17:47











  • Possible duplicate of How do you construct a std::string with an embedded null?

    – πάντα ῥεῖ
    Mar 6 at 18:00















3















When running this code I get "Hello Worldaaa", but according to the job of the .c_str() function, the string a should be "Hello World" and string b should be "Hello Worldaaa" so "aaa" should not show up in the output.



#include <bits/stdc++.h>

using namespace std;

int main()

string a = "Hello World";
a = a.c_str();
string b = a + string("aaa");

cout << b;

return 0;










share|improve this question

















  • 3





    "Returns a pointer to a null-terminated character array with data equivalent to those stored in the string." It does, but using it in the std::string operator+(const char*, const std::string&) ignores the terminating '' character to produce the result.

    – πάντα ῥεῖ
    Mar 6 at 17:46






  • 2





    it's not actually .c_str(), it's actually the constructor that's confusing you.

    – Mooing Duck
    Mar 6 at 17:47











  • Possible duplicate of How do you construct a std::string with an embedded null?

    – πάντα ῥεῖ
    Mar 6 at 18:00













3












3








3








When running this code I get "Hello Worldaaa", but according to the job of the .c_str() function, the string a should be "Hello World" and string b should be "Hello Worldaaa" so "aaa" should not show up in the output.



#include <bits/stdc++.h>

using namespace std;

int main()

string a = "Hello World";
a = a.c_str();
string b = a + string("aaa");

cout << b;

return 0;










share|improve this question














When running this code I get "Hello Worldaaa", but according to the job of the .c_str() function, the string a should be "Hello World" and string b should be "Hello Worldaaa" so "aaa" should not show up in the output.



#include <bits/stdc++.h>

using namespace std;

int main()

string a = "Hello World";
a = a.c_str();
string b = a + string("aaa");

cout << b;

return 0;







c++ c-strings






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Mar 6 at 17:42









Mayar KaroutMayar Karout

293




293







  • 3





    "Returns a pointer to a null-terminated character array with data equivalent to those stored in the string." It does, but using it in the std::string operator+(const char*, const std::string&) ignores the terminating '' character to produce the result.

    – πάντα ῥεῖ
    Mar 6 at 17:46






  • 2





    it's not actually .c_str(), it's actually the constructor that's confusing you.

    – Mooing Duck
    Mar 6 at 17:47











  • Possible duplicate of How do you construct a std::string with an embedded null?

    – πάντα ῥεῖ
    Mar 6 at 18:00












  • 3





    "Returns a pointer to a null-terminated character array with data equivalent to those stored in the string." It does, but using it in the std::string operator+(const char*, const std::string&) ignores the terminating '' character to produce the result.

    – πάντα ῥεῖ
    Mar 6 at 17:46






  • 2





    it's not actually .c_str(), it's actually the constructor that's confusing you.

    – Mooing Duck
    Mar 6 at 17:47











  • Possible duplicate of How do you construct a std::string with an embedded null?

    – πάντα ῥεῖ
    Mar 6 at 18:00







3




3





"Returns a pointer to a null-terminated character array with data equivalent to those stored in the string." It does, but using it in the std::string operator+(const char*, const std::string&) ignores the terminating '' character to produce the result.

– πάντα ῥεῖ
Mar 6 at 17:46





"Returns a pointer to a null-terminated character array with data equivalent to those stored in the string." It does, but using it in the std::string operator+(const char*, const std::string&) ignores the terminating '' character to produce the result.

– πάντα ῥεῖ
Mar 6 at 17:46




2




2





it's not actually .c_str(), it's actually the constructor that's confusing you.

– Mooing Duck
Mar 6 at 17:47





it's not actually .c_str(), it's actually the constructor that's confusing you.

– Mooing Duck
Mar 6 at 17:47













Possible duplicate of How do you construct a std::string with an embedded null?

– πάντα ῥεῖ
Mar 6 at 18:00





Possible duplicate of How do you construct a std::string with an embedded null?

– πάντα ῥεῖ
Mar 6 at 18:00












3 Answers
3






active

oldest

votes


















4














Yes, you are right regarding
std::string::c_str.



Indeed,




Returns a pointer to a null-terminated character array with data equivalent to those stored in the string.




However, there are a few considerations which probably misleading you.




First of all, the statement:



a = a.c_str();


is semantically a no-operation.
You get the const char* (i.e. "Hello World") and assign it to a.
However a is a std::string, it is a class designed to abstract a "string type" in C++. It handles automatically the '', transparently. The user is not supposed to care about the managing of it1.



The trick is inside std::string::operator=(const char*).




Replaces the contents with those of null-terminated character string pointed to by s as if by assign(s, Traits::length(s)).





The last point, about string concatenation.



a + string("aaa");


As previously, in this case, the operator std::string::operator+ will handle the ''.



Shortly, it behaves like:



"Hello World" + "aaa" ===> "Hello Worldaaa"


It will care about the "inner ''" returning a conforming null-terminated string.




1 Unless he/she is playing with the internal memory.






share|improve this answer























  • Really, the point needs to be made that in a "NUL-terminated string", the NUL terminator is not considered part of the string's data. It is merely an end-of-string indicator. That is, "Hello World" is a different string from "Hello World"; the latter has an embedded NUL character which is part of the actual string.

    – Nicol Bolas
    Mar 6 at 18:24


















1














That would be unintuitive. Remember that the terminating null character wasn't explicitly used by the user. Syntax of the language added it.



It will create a lot of confusion if the result of "abcd" + "xyz" ends up being "abcdxyz" instead of the intuitive result - "abcdxyz".



In C and C++, concatenating two strings implies ignoring the terminating null character of the first string in the resultant string. Take a look at the documentation of strcat. It is explicit about the null character while the documentation of std::string::operator+ isn't.






share|improve this answer

























  • The really interesting part is what would be the workaround to store the character explicitely in a std::string.

    – πάντα ῥεῖ
    Mar 6 at 17:52











  • @πάνταῥεῖ, I am sure people with a more computer science/programming language theory bent of mind have thought about it. Not me.

    – R Sahu
    Mar 6 at 17:57






  • 1





    Yes they did

    – πάντα ῥεῖ
    Mar 6 at 17:59


















0














Not quite...



Let's take a look at this line:



a = a.c_str();


You have an std::string that you're trying to assign a const char* to. So you'll be using overload (3) of the std::string assignment operator. But this doesn't magically change the type of a. In other words, we still have an std::string!



So that line doesn't actually modify the value of a at all.



Then you use the operator+ to add two std::string's together. This will result in another std::string that is the first with the second appended, which is exactly what you have!



It's not clear to me what you're trying to accomplish here, but if you created a c-string with those values:



const char cstr[] = "Hello Worldaaa";


And tried to print it, you'd see exactly what you're expecting.






share|improve this answer
























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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    Yes, you are right regarding
    std::string::c_str.



    Indeed,




    Returns a pointer to a null-terminated character array with data equivalent to those stored in the string.




    However, there are a few considerations which probably misleading you.




    First of all, the statement:



    a = a.c_str();


    is semantically a no-operation.
    You get the const char* (i.e. "Hello World") and assign it to a.
    However a is a std::string, it is a class designed to abstract a "string type" in C++. It handles automatically the '', transparently. The user is not supposed to care about the managing of it1.



    The trick is inside std::string::operator=(const char*).




    Replaces the contents with those of null-terminated character string pointed to by s as if by assign(s, Traits::length(s)).





    The last point, about string concatenation.



    a + string("aaa");


    As previously, in this case, the operator std::string::operator+ will handle the ''.



    Shortly, it behaves like:



    "Hello World" + "aaa" ===> "Hello Worldaaa"


    It will care about the "inner ''" returning a conforming null-terminated string.




    1 Unless he/she is playing with the internal memory.






    share|improve this answer























    • Really, the point needs to be made that in a "NUL-terminated string", the NUL terminator is not considered part of the string's data. It is merely an end-of-string indicator. That is, "Hello World" is a different string from "Hello World"; the latter has an embedded NUL character which is part of the actual string.

      – Nicol Bolas
      Mar 6 at 18:24















    4














    Yes, you are right regarding
    std::string::c_str.



    Indeed,




    Returns a pointer to a null-terminated character array with data equivalent to those stored in the string.




    However, there are a few considerations which probably misleading you.




    First of all, the statement:



    a = a.c_str();


    is semantically a no-operation.
    You get the const char* (i.e. "Hello World") and assign it to a.
    However a is a std::string, it is a class designed to abstract a "string type" in C++. It handles automatically the '', transparently. The user is not supposed to care about the managing of it1.



    The trick is inside std::string::operator=(const char*).




    Replaces the contents with those of null-terminated character string pointed to by s as if by assign(s, Traits::length(s)).





    The last point, about string concatenation.



    a + string("aaa");


    As previously, in this case, the operator std::string::operator+ will handle the ''.



    Shortly, it behaves like:



    "Hello World" + "aaa" ===> "Hello Worldaaa"


    It will care about the "inner ''" returning a conforming null-terminated string.




    1 Unless he/she is playing with the internal memory.






    share|improve this answer























    • Really, the point needs to be made that in a "NUL-terminated string", the NUL terminator is not considered part of the string's data. It is merely an end-of-string indicator. That is, "Hello World" is a different string from "Hello World"; the latter has an embedded NUL character which is part of the actual string.

      – Nicol Bolas
      Mar 6 at 18:24













    4












    4








    4







    Yes, you are right regarding
    std::string::c_str.



    Indeed,




    Returns a pointer to a null-terminated character array with data equivalent to those stored in the string.




    However, there are a few considerations which probably misleading you.




    First of all, the statement:



    a = a.c_str();


    is semantically a no-operation.
    You get the const char* (i.e. "Hello World") and assign it to a.
    However a is a std::string, it is a class designed to abstract a "string type" in C++. It handles automatically the '', transparently. The user is not supposed to care about the managing of it1.



    The trick is inside std::string::operator=(const char*).




    Replaces the contents with those of null-terminated character string pointed to by s as if by assign(s, Traits::length(s)).





    The last point, about string concatenation.



    a + string("aaa");


    As previously, in this case, the operator std::string::operator+ will handle the ''.



    Shortly, it behaves like:



    "Hello World" + "aaa" ===> "Hello Worldaaa"


    It will care about the "inner ''" returning a conforming null-terminated string.




    1 Unless he/she is playing with the internal memory.






    share|improve this answer













    Yes, you are right regarding
    std::string::c_str.



    Indeed,




    Returns a pointer to a null-terminated character array with data equivalent to those stored in the string.




    However, there are a few considerations which probably misleading you.




    First of all, the statement:



    a = a.c_str();


    is semantically a no-operation.
    You get the const char* (i.e. "Hello World") and assign it to a.
    However a is a std::string, it is a class designed to abstract a "string type" in C++. It handles automatically the '', transparently. The user is not supposed to care about the managing of it1.



    The trick is inside std::string::operator=(const char*).




    Replaces the contents with those of null-terminated character string pointed to by s as if by assign(s, Traits::length(s)).





    The last point, about string concatenation.



    a + string("aaa");


    As previously, in this case, the operator std::string::operator+ will handle the ''.



    Shortly, it behaves like:



    "Hello World" + "aaa" ===> "Hello Worldaaa"


    It will care about the "inner ''" returning a conforming null-terminated string.




    1 Unless he/she is playing with the internal memory.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Mar 6 at 17:56









    Biagio FestaBiagio Festa

    5,11821239




    5,11821239












    • Really, the point needs to be made that in a "NUL-terminated string", the NUL terminator is not considered part of the string's data. It is merely an end-of-string indicator. That is, "Hello World" is a different string from "Hello World"; the latter has an embedded NUL character which is part of the actual string.

      – Nicol Bolas
      Mar 6 at 18:24

















    • Really, the point needs to be made that in a "NUL-terminated string", the NUL terminator is not considered part of the string's data. It is merely an end-of-string indicator. That is, "Hello World" is a different string from "Hello World"; the latter has an embedded NUL character which is part of the actual string.

      – Nicol Bolas
      Mar 6 at 18:24
















    Really, the point needs to be made that in a "NUL-terminated string", the NUL terminator is not considered part of the string's data. It is merely an end-of-string indicator. That is, "Hello World" is a different string from "Hello World"; the latter has an embedded NUL character which is part of the actual string.

    – Nicol Bolas
    Mar 6 at 18:24





    Really, the point needs to be made that in a "NUL-terminated string", the NUL terminator is not considered part of the string's data. It is merely an end-of-string indicator. That is, "Hello World" is a different string from "Hello World"; the latter has an embedded NUL character which is part of the actual string.

    – Nicol Bolas
    Mar 6 at 18:24













    1














    That would be unintuitive. Remember that the terminating null character wasn't explicitly used by the user. Syntax of the language added it.



    It will create a lot of confusion if the result of "abcd" + "xyz" ends up being "abcdxyz" instead of the intuitive result - "abcdxyz".



    In C and C++, concatenating two strings implies ignoring the terminating null character of the first string in the resultant string. Take a look at the documentation of strcat. It is explicit about the null character while the documentation of std::string::operator+ isn't.






    share|improve this answer

























    • The really interesting part is what would be the workaround to store the character explicitely in a std::string.

      – πάντα ῥεῖ
      Mar 6 at 17:52











    • @πάνταῥεῖ, I am sure people with a more computer science/programming language theory bent of mind have thought about it. Not me.

      – R Sahu
      Mar 6 at 17:57






    • 1





      Yes they did

      – πάντα ῥεῖ
      Mar 6 at 17:59















    1














    That would be unintuitive. Remember that the terminating null character wasn't explicitly used by the user. Syntax of the language added it.



    It will create a lot of confusion if the result of "abcd" + "xyz" ends up being "abcdxyz" instead of the intuitive result - "abcdxyz".



    In C and C++, concatenating two strings implies ignoring the terminating null character of the first string in the resultant string. Take a look at the documentation of strcat. It is explicit about the null character while the documentation of std::string::operator+ isn't.






    share|improve this answer

























    • The really interesting part is what would be the workaround to store the character explicitely in a std::string.

      – πάντα ῥεῖ
      Mar 6 at 17:52











    • @πάνταῥεῖ, I am sure people with a more computer science/programming language theory bent of mind have thought about it. Not me.

      – R Sahu
      Mar 6 at 17:57






    • 1





      Yes they did

      – πάντα ῥεῖ
      Mar 6 at 17:59













    1












    1








    1







    That would be unintuitive. Remember that the terminating null character wasn't explicitly used by the user. Syntax of the language added it.



    It will create a lot of confusion if the result of "abcd" + "xyz" ends up being "abcdxyz" instead of the intuitive result - "abcdxyz".



    In C and C++, concatenating two strings implies ignoring the terminating null character of the first string in the resultant string. Take a look at the documentation of strcat. It is explicit about the null character while the documentation of std::string::operator+ isn't.






    share|improve this answer















    That would be unintuitive. Remember that the terminating null character wasn't explicitly used by the user. Syntax of the language added it.



    It will create a lot of confusion if the result of "abcd" + "xyz" ends up being "abcdxyz" instead of the intuitive result - "abcdxyz".



    In C and C++, concatenating two strings implies ignoring the terminating null character of the first string in the resultant string. Take a look at the documentation of strcat. It is explicit about the null character while the documentation of std::string::operator+ isn't.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Mar 6 at 17:59

























    answered Mar 6 at 17:49









    R SahuR Sahu

    168k1292191




    168k1292191












    • The really interesting part is what would be the workaround to store the character explicitely in a std::string.

      – πάντα ῥεῖ
      Mar 6 at 17:52











    • @πάνταῥεῖ, I am sure people with a more computer science/programming language theory bent of mind have thought about it. Not me.

      – R Sahu
      Mar 6 at 17:57






    • 1





      Yes they did

      – πάντα ῥεῖ
      Mar 6 at 17:59

















    • The really interesting part is what would be the workaround to store the character explicitely in a std::string.

      – πάντα ῥεῖ
      Mar 6 at 17:52











    • @πάνταῥεῖ, I am sure people with a more computer science/programming language theory bent of mind have thought about it. Not me.

      – R Sahu
      Mar 6 at 17:57






    • 1





      Yes they did

      – πάντα ῥεῖ
      Mar 6 at 17:59
















    The really interesting part is what would be the workaround to store the character explicitely in a std::string.

    – πάντα ῥεῖ
    Mar 6 at 17:52





    The really interesting part is what would be the workaround to store the character explicitely in a std::string.

    – πάντα ῥεῖ
    Mar 6 at 17:52













    @πάνταῥεῖ, I am sure people with a more computer science/programming language theory bent of mind have thought about it. Not me.

    – R Sahu
    Mar 6 at 17:57





    @πάνταῥεῖ, I am sure people with a more computer science/programming language theory bent of mind have thought about it. Not me.

    – R Sahu
    Mar 6 at 17:57




    1




    1





    Yes they did

    – πάντα ῥεῖ
    Mar 6 at 17:59





    Yes they did

    – πάντα ῥεῖ
    Mar 6 at 17:59











    0














    Not quite...



    Let's take a look at this line:



    a = a.c_str();


    You have an std::string that you're trying to assign a const char* to. So you'll be using overload (3) of the std::string assignment operator. But this doesn't magically change the type of a. In other words, we still have an std::string!



    So that line doesn't actually modify the value of a at all.



    Then you use the operator+ to add two std::string's together. This will result in another std::string that is the first with the second appended, which is exactly what you have!



    It's not clear to me what you're trying to accomplish here, but if you created a c-string with those values:



    const char cstr[] = "Hello Worldaaa";


    And tried to print it, you'd see exactly what you're expecting.






    share|improve this answer





























      0














      Not quite...



      Let's take a look at this line:



      a = a.c_str();


      You have an std::string that you're trying to assign a const char* to. So you'll be using overload (3) of the std::string assignment operator. But this doesn't magically change the type of a. In other words, we still have an std::string!



      So that line doesn't actually modify the value of a at all.



      Then you use the operator+ to add two std::string's together. This will result in another std::string that is the first with the second appended, which is exactly what you have!



      It's not clear to me what you're trying to accomplish here, but if you created a c-string with those values:



      const char cstr[] = "Hello Worldaaa";


      And tried to print it, you'd see exactly what you're expecting.






      share|improve this answer



























        0












        0








        0







        Not quite...



        Let's take a look at this line:



        a = a.c_str();


        You have an std::string that you're trying to assign a const char* to. So you'll be using overload (3) of the std::string assignment operator. But this doesn't magically change the type of a. In other words, we still have an std::string!



        So that line doesn't actually modify the value of a at all.



        Then you use the operator+ to add two std::string's together. This will result in another std::string that is the first with the second appended, which is exactly what you have!



        It's not clear to me what you're trying to accomplish here, but if you created a c-string with those values:



        const char cstr[] = "Hello Worldaaa";


        And tried to print it, you'd see exactly what you're expecting.






        share|improve this answer















        Not quite...



        Let's take a look at this line:



        a = a.c_str();


        You have an std::string that you're trying to assign a const char* to. So you'll be using overload (3) of the std::string assignment operator. But this doesn't magically change the type of a. In other words, we still have an std::string!



        So that line doesn't actually modify the value of a at all.



        Then you use the operator+ to add two std::string's together. This will result in another std::string that is the first with the second appended, which is exactly what you have!



        It's not clear to me what you're trying to accomplish here, but if you created a c-string with those values:



        const char cstr[] = "Hello Worldaaa";


        And tried to print it, you'd see exactly what you're expecting.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Mar 6 at 17:53

























        answered Mar 6 at 17:50









        scohe001scohe001

        8,15212342




        8,15212342



























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