printf with std string

Question

My understanding is that string is a member of the std namespace  so why does the following occur    include  lt iostream gt   int main         using namespace std       string myString    Press ENTER to quit program        cout  lt  lt   Come up and C   me some time    lt  lt  endl      printf  Follow this command   s   myString       cin get         return 0        Each time the program runs  myString prints a seemingly random string of 3 characters  such as in the output above

User · Answer

Printf is actually pretty good to use if size matters. Meaning if you are running a program where memory is an issue, then printf is actually a very good and under rater solution. Cout essentially shifts bits over to make room for the string, while printf just takes in some sort of parameters and prints it to the screen. If you were to compile a simple hello world program, printf would be able to compile it in less than 60, 000 bits as opposed to cout, it would take over 1 million bits to compile.

For your situation, id suggest using cout simply because it is much more convenient to use. Although, I would argue that printf is something good to know.

User · Answer

Please don t use printf   s   your string c str      Use cout  lt  lt  your string  instead  Short  simple and typesafe  In fact  when you re writing C    you generally want to avoid printf entirely -- it s a leftover from C that s rarely needed or useful in C     As to why you should use cout instead of printf  the reasons are numerous  Here s a sampling of a few of the most obvious    As the question shows  printf isn t type-safe  If the type you pass differs from that given in the conversion specifier  printf will try to use whatever it finds on the stack as if it were the specified type  giving undefined behavior  Some compilers can warn about this under some circumstances  but some compilers can t won t at all  and none can under all circumstances  printf isn t extensible  You can only pass primitive types to it  The set of conversion specifiers it understands is hard-coded in its implementation  and there s no way for you to add more others  Most well-written C   should use these types primarily to implement types oriented toward the problem being solved  It makes decent formatting much more difficult  For an obvious example  when you re printing numbers for people to read  you typically want to insert thousands separators every few digits  The exact number of digits and the characters used as separators varies  but cout has that covered as well  For example   std  locale loc      std  cout imbue loc    std  cout  lt  lt  123456 78    The nameless locale  the     picks a locale based on the user s configuration  Therefore  on my machine  configured for US English  this prints out as 123 456 78  For somebody who has their computer configured for  say  Germany  it would print out something like 123 456 78  For somebody with it configured for India  it would print out as 1 23 456 78  and of course there are many others   With printf I get exactly one result  123456 78  It is consistent  but it s consistently wrong for everybody everywhere  Essentially the only way to work around it is to do the formatting separately  then pass the result as a string to printf  because printf itself simply will not do the job correctly  Although they re quite compact  printf format strings can be quite unreadable  Even among C programmers who use printf virtually every day  I d guess at least 99  would need to look things up to be sure what the   in   x means  and how that differs from what the   in   f means  and yes  they mean entirely different things

User · Answer

Use std  printf and c str   example    std  printf  Follow this command   s   myString c str

User · Answer

printf accepts a variable number of arguments  Those can only have Plain Old Data  POD  types  Code that passes anything other than POD to printf only compiles because the compiler assumes you got your format right   s means that the respective argument is supposed to be a pointer to a char  In your case it is an std  string not const char   printf does not know it because the argument type goes lost and is supposed to be restored from the format parameter  When turning that std  string argument into const char  the resulting pointer will point to some irrelevant region of memory instead of your desired C string  For that reason your code prints out gibberish   While printf is an excellent choice for printing out formatted text   especially if you intend to have padding   it can be dangerous if you haven t enabled compiler warnings  Always enable warnings because then mistakes like this are easily avoidable  There is no reason to use the clumsy std  cout mechanism if the printf family can do the same task in a much faster and prettier way  Just make sure you have enabled all warnings  -Wall -Wextra  and you will be good  In case you use your own custom printf implementation you should declare it with the   attribute   mechanism that enables the compiler to check the format string against the parameters provided

User · Answer

It s compiling because printf isn t type safe  since it uses variable arguments in the C sense1  printf has no option for std  string  only a C-style string  Using something else in place of what it expects definitely won t give you the results you want  It s actually undefined behaviour  so anything at all could happen   The easiest way to fix this  since you re using C    is printing it normally with std  cout  since std  string supports that through operator overloading   std  cout  lt  lt   Follow this command     lt  lt  myString    If  for some reason  you need to extract the C-style string  you can use the c str   method of std  string to get a const char   that is null-terminated  Using your example    include  lt iostream gt   include  lt string gt   include  lt stdio h gt   int main         using namespace std       string myString    Press ENTER to quit program        cout  lt  lt   Come up and C   me some time    lt  lt  endl      printf  Follow this command   s   myString c str       note the use of c str     cin get         return 0      If you want a function that is like printf  but type safe  look into variadic templates  C  11  supported on all major compilers as of MSVC12   You can find an example of one here  There s nothing I know of implemented like that in the standard library  but there might be in Boost  specifically boost  format      1   This means that you can pass any number of arguments  but the function relies on you to tell it the number and types of those arguments  In the case of printf  that means a string with encoded type information like  d meaning int  If you lie about the type or number  the function has no standard way of knowing  although some compilers have the ability to check and give warnings when you lie

User · Answer

The main reason is probably that a C   string is a struct that includes a current-length value  not just the address of a sequence of chars terminated by a 0 byte   Printf and its relatives expect to find such a sequence  not a struct  and therefore get confused by C   strings   Speaking for myself  I believe that printf has a place that can t easily be filled by C   syntactic features  just as table structures in html have a place that can t easily be filled by divs   As Dykstra wrote later about the goto  he didn t intend to start a religion and was really only arguing against using it as a kludge to make up for poorly-designed code     It would be quite nice if the GNU project would add the printf family to their g   extensions

User · Answer

use myString c str   if you want a c-like string  const char   to use with printf  thanks

[c++] printf with std::string?

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