How do you append an int to a string in C

Question

int i   4  string text    Player    cout  lt  lt   text   i     I d like it to print Player 4   The above is obviously wrong but it shows what I m trying to do here  Is there an easy way to do this or do I have to start adding new includes

User · Answer

For the record  you can also use a std  stringstream if you want to create the string before it s actually output

User · Answer

If using Windows MFC  and need the string for more than immediate output try   int i   4  CString strOutput  strOutput Format  Player  d   i

User · Answer

These work for general strings  in case you do not want to output to file console  but store for later use or something    boost lexical cast  MyStr    boost  lexical cast lt std  string gt  MyInt     String streams    sstream h std  stringstream Stream  Stream str MyStr   Stream  lt  lt  MyInt  MyStr   Stream str        If you re using a stream  for example  cout   rather than std  string someStream  lt  lt  MyInt

User · Answer

These work for general strings  in case you do not want to output to file console  but store for later use or something    boost lexical cast  MyStr    boost  lexical cast lt std  string gt  MyInt     String streams    sstream h std  stringstream Stream  Stream str MyStr   Stream  lt  lt  MyInt  MyStr   Stream str        If you re using a stream  for example  cout   rather than std  string someStream  lt  lt  MyInt

User · Answer

Another possibility is Boost Format    include  lt boost format hpp gt   include  lt iostream gt   include  lt string gt   int main       int i   4    std  string text    Player     std  cout  lt  lt  boost  format   1   2  n     text   i

User · Answer

For the record  you could also use Qt s QString class    include  lt QtCore QString gt   int i   4  QString qs   QString  Player  1   arg i   std  cout  lt  lt  qs toLocal8bit   constData        prints  Player 4

User · Answer

Well  if you use cout you can just write the integer directly to it  as in  std  cout  lt  lt  text  lt  lt  i    The C   way of converting all kinds of objects to strings is through string streams  If you don t have one handy  just create one    include  lt sstream gt   std  ostringstream oss  oss  lt  lt  text  lt  lt  i  std  cout  lt  lt  oss str      Alternatively  you can just convert the integer and append it to the string   oss  lt  lt  i  text    oss str      Finally  the Boost libraries provide boost  lexical cast  which wraps around the stringstream conversion with a syntax like the built-in type casts    include  lt boost lexical cast hpp gt   text    boost  lexical cast lt std  string gt  i     This also works the other way around  i e  to parse strings

User · Answer

For the record  you can also use a std  stringstream if you want to create the string before it s actually output

User · Answer

Your example seems to indicate that you would like to display the a string followed by an integer  in which case   string text    Player     int i   4  cout  lt  lt  text  lt  lt  i  lt  lt  endl    would work fine   But  if you re going to be storing the string places or passing it around  and doing this frequently  you may benefit from overloading the addition operator  I demonstrate this below    include  lt sstream gt   include  lt iostream gt  using namespace std   std  string operator  std  string const  amp a  int b      std  ostringstream oss    oss  lt  lt  a  lt  lt  b    return oss str       int main       int i   4    string text    Player       cout  lt  lt   text   i   lt  lt  endl      In fact  you can use templates to make this approach more powerful   template  lt class T gt  std  string operator  std  string const  amp a  const T  amp b     std  ostringstream oss    oss  lt  lt  a  lt  lt  b    return oss str        Now  as long as object b has a defined stream output  you can append it to your string  or  at least  a copy thereof

User · Answer

You also try concatenate player s number with std  string  push back    Example with your code   int i   4  string text    Player    text push back i    0    cout  lt  lt  text    You will see in console       Player 4

User · Answer

For the record  you can also use a std  stringstream if you want to create the string before it s actually output

User · Answer

These work for general strings  in case you do not want to output to file console  but store for later use or something    boost lexical cast  MyStr    boost  lexical cast lt std  string gt  MyInt     String streams    sstream h std  stringstream Stream  Stream str MyStr   Stream  lt  lt  MyInt  MyStr   Stream str        If you re using a stream  for example  cout   rather than std  string someStream  lt  lt  MyInt

User · Answer

cout  lt  lt  text  lt  lt       lt  lt  i  lt  lt  endl

User · Answer

cout  lt  lt   Player   lt  lt  i

User · Answer

cout  lt  lt   Player   lt  lt  i

User · Answer

cout  lt  lt  text  lt  lt       lt  lt  i  lt  lt  endl

User · Answer

Here a small working conversion appending example  with some code I needed before    include  lt string gt   include  lt sstream gt   include  lt iostream gt   using namespace std   int main    string str  int i   321  std  stringstream ss  ss  lt  lt  123  str     dev video   cout  lt  lt  str  lt  lt  endl  cout  lt  lt  str  lt  lt  456  lt  lt  endl  cout  lt  lt  str  lt  lt  i  lt  lt  endl  str    ss str    cout  lt  lt  str  lt  lt  endl      the output will be    dev video  dev video456  dev video321  dev video123   Note that in the last two lines you save the modified string before it s actually printed out  and you could use it later if needed

User · Answer

You can use the following  int i   4  string text    Player    text   i  0    cout  lt  lt   text

User · Answer

cout  lt  lt  text  lt  lt  i

User · Answer

printf  Player  d   i      Downvote my answer all you like  I still hate the C   I O operators     -P

User · Answer

cout  lt  lt  text  lt  lt       lt  lt  i  lt  lt  endl

User · Answer

You can use the following  int i   4  string text    Player    text   i  0    cout  lt  lt   text

User · Answer

For the record  you can also use a std  stringstream if you want to create the string before it s actually output

User · Answer

Your example seems to indicate that you would like to display the a string followed by an integer  in which case   string text    Player     int i   4  cout  lt  lt  text  lt  lt  i  lt  lt  endl    would work fine   But  if you re going to be storing the string places or passing it around  and doing this frequently  you may benefit from overloading the addition operator  I demonstrate this below    include  lt sstream gt   include  lt iostream gt  using namespace std   std  string operator  std  string const  amp a  int b      std  ostringstream oss    oss  lt  lt  a  lt  lt  b    return oss str       int main       int i   4    string text    Player       cout  lt  lt   text   i   lt  lt  endl      In fact  you can use templates to make this approach more powerful   template  lt class T gt  std  string operator  std  string const  amp a  const T  amp b     std  ostringstream oss    oss  lt  lt  a  lt  lt  b    return oss str        Now  as long as object b has a defined stream output  you can append it to your string  or  at least  a copy thereof

User · Answer

cout  lt  lt  text  lt  lt  i    The  lt  lt  operator for ostream returns a reference to the ostream  so you can just keep chaining the  lt  lt  operations   That is  the above is basically the same as   cout  lt  lt  text  cout  lt  lt  i

User · Answer

cout  lt  lt  text  lt  lt       lt  lt  i  lt  lt  endl

User · Answer

You also try concatenate player s number with std  string  push back    Example with your code   int i   4  string text    Player    text push back i    0    cout  lt  lt  text    You will see in console       Player 4

User · Answer

There are a few options  and which one you want depends on the context   The simplest way is  std  cout  lt  lt  text  lt  lt  i    or if you want this on a single line  std  cout  lt  lt  text  lt  lt  i  lt  lt  endl    If you are writing a single threaded program and if you aren t calling this code a lot  where  a lot  is thousands of times per second  then you are done   If you are writing a multi threaded program and more than one thread is writing to cout  then this simple code can get you into trouble  Let s assume that the library that came with your compiler made cout thread safe enough than any single call to it won t be interrupted  Now let s say that one thread is using this code to write  Player 1  and another is writing  Player 2   If you are lucky you will get the following   Player 1 Player 2   If you are unlucky you might get something like the following  Player Player 2 1   The problem is that std  cout  lt  lt  text  lt  lt  i  lt  lt  endl  turns into 3 function calls  The code is equivalent to the following   std  cout  lt  lt  text  std  cout  lt  lt  i  std  cout  lt  lt  endl    If instead you used the C-style printf  and again your compiler provided a runtime library with reasonable thread safety  each function call is atomic  then the following code would work better   printf  Player  d n   i     Being able to do something in a single function call lets the io library provide synchronization under the covers  and now your whole line of text will be atomically written   For simple programs  std  cout is great  Throw in multithreading or other complications and the less stylish printf starts to look more attractive

User · Answer

For the record  you could also use Qt s QString class    include  lt QtCore QString gt   int i   4  QString qs   QString  Player  1   arg i   std  cout  lt  lt  qs toLocal8bit   constData        prints  Player 4

User · Answer

cout  lt  lt  text  lt  lt  i

User · Answer

With C  11  you can write    include  lt string gt         to use std  string  std  to string   and     operator acting on strings   int i   4  std  string text    Player    text    std  to string i

User · Answer

printf  Player  d   i      Downvote my answer all you like  I still hate the C   I O operators     -P

User · Answer

Well  if you use cout you can just write the integer directly to it  as in  std  cout  lt  lt  text  lt  lt  i    The C   way of converting all kinds of objects to strings is through string streams  If you don t have one handy  just create one    include  lt sstream gt   std  ostringstream oss  oss  lt  lt  text  lt  lt  i  std  cout  lt  lt  oss str      Alternatively  you can just convert the integer and append it to the string   oss  lt  lt  i  text    oss str      Finally  the Boost libraries provide boost  lexical cast  which wraps around the stringstream conversion with a syntax like the built-in type casts    include  lt boost lexical cast hpp gt   text    boost  lexical cast lt std  string gt  i     This also works the other way around  i e  to parse strings

User · Answer

The easiest way I could figure this out is the following   It will work as a single string and string array  I am considering a string array  as it is complicated  little bit same will be followed with string   I create a array of names and append some integer and char with it to show how easy it is to append some int and chars to string  hope it helps  length is just to measure the size of array  If you are familiar with programming then size t is a unsigned int   include lt iostream gt       include lt string gt      using namespace std      int main              string names        amz   Waq   Mon   Sam   Has   Shak   GBy       simple array         int length   sizeof names    sizeof names 0      give you size of array         int id          string append 7        as length is 7 just for sake of storing and printing output          for  size t i   0  i  lt  length  i                  id   rand     20000   2              append i    names i    to string id                     for  size t i   0  i  lt  length  i                  cout  lt  lt  append i   lt  lt  endl

User · Answer

cout  lt  lt  text  lt  lt       lt  lt  i  lt  lt  endl

User · Answer

cout  lt  lt  text  lt  lt       lt  lt  i  lt  lt  endl

User · Answer

There are a few options  and which one you want depends on the context   The simplest way is  std  cout  lt  lt  text  lt  lt  i    or if you want this on a single line  std  cout  lt  lt  text  lt  lt  i  lt  lt  endl    If you are writing a single threaded program and if you aren t calling this code a lot  where  a lot  is thousands of times per second  then you are done   If you are writing a multi threaded program and more than one thread is writing to cout  then this simple code can get you into trouble  Let s assume that the library that came with your compiler made cout thread safe enough than any single call to it won t be interrupted  Now let s say that one thread is using this code to write  Player 1  and another is writing  Player 2   If you are lucky you will get the following   Player 1 Player 2   If you are unlucky you might get something like the following  Player Player 2 1   The problem is that std  cout  lt  lt  text  lt  lt  i  lt  lt  endl  turns into 3 function calls  The code is equivalent to the following   std  cout  lt  lt  text  std  cout  lt  lt  i  std  cout  lt  lt  endl    If instead you used the C-style printf  and again your compiler provided a runtime library with reasonable thread safety  each function call is atomic  then the following code would work better   printf  Player  d n   i     Being able to do something in a single function call lets the io library provide synchronization under the covers  and now your whole line of text will be atomically written   For simple programs  std  cout is great  Throw in multithreading or other complications and the less stylish printf starts to look more attractive

User · Answer

Here a small working conversion appending example  with some code I needed before    include  lt string gt   include  lt sstream gt   include  lt iostream gt   using namespace std   int main    string str  int i   321  std  stringstream ss  ss  lt  lt  123  str     dev video   cout  lt  lt  str  lt  lt  endl  cout  lt  lt  str  lt  lt  456  lt  lt  endl  cout  lt  lt  str  lt  lt  i  lt  lt  endl  str    ss str    cout  lt  lt  str  lt  lt  endl      the output will be    dev video  dev video456  dev video321  dev video123   Note that in the last two lines you save the modified string before it s actually printed out  and you could use it later if needed

User · Answer

These work for general strings  in case you do not want to output to file console  but store for later use or something    boost lexical cast  MyStr    boost  lexical cast lt std  string gt  MyInt     String streams    sstream h std  stringstream Stream  Stream str MyStr   Stream  lt  lt  MyInt  MyStr   Stream str        If you re using a stream  for example  cout   rather than std  string someStream  lt  lt  MyInt

User · Answer

printf  Player  d   i      Downvote my answer all you like  I still hate the C   I O operators     -P

User · Answer

Well  if you use cout you can just write the integer directly to it  as in  std  cout  lt  lt  text  lt  lt  i    The C   way of converting all kinds of objects to strings is through string streams  If you don t have one handy  just create one    include  lt sstream gt   std  ostringstream oss  oss  lt  lt  text  lt  lt  i  std  cout  lt  lt  oss str      Alternatively  you can just convert the integer and append it to the string   oss  lt  lt  i  text    oss str      Finally  the Boost libraries provide boost  lexical cast  which wraps around the stringstream conversion with a syntax like the built-in type casts    include  lt boost lexical cast hpp gt   text    boost  lexical cast lt std  string gt  i     This also works the other way around  i e  to parse strings

User · Answer

Another possibility is Boost Format    include  lt boost format hpp gt   include  lt iostream gt   include  lt string gt   int main       int i   4    std  string text    Player     std  cout  lt  lt  boost  format   1   2  n     text   i

User · Answer

One method here is directly printing the output if its required in your problem   cout  lt  lt  text  lt  lt  i    Else  one of the safest method is to use   sprintf count    d   i     And then copy it to your  text  string    for k   0    count   k   k         text    count k         Thus  you have your required output string   For more info on sprintf  follow  http   www cplusplus com reference cstdio sprintf

User · Answer

With C  11  you can write    include  lt string gt         to use std  string  std  to string   and     operator acting on strings   int i   4  std  string text    Player    text    std  to string i

User · Answer

One method here is directly printing the output if its required in your problem   cout  lt  lt  text  lt  lt  i    Else  one of the safest method is to use   sprintf count    d   i     And then copy it to your  text  string    for k   0    count   k   k         text    count k         Thus  you have your required output string   For more info on sprintf  follow  http   www cplusplus com reference cstdio sprintf

User · Answer

The easiest way I could figure this out is the following   It will work as a single string and string array  I am considering a string array  as it is complicated  little bit same will be followed with string   I create a array of names and append some integer and char with it to show how easy it is to append some int and chars to string  hope it helps  length is just to measure the size of array  If you are familiar with programming then size t is a unsigned int   include lt iostream gt       include lt string gt      using namespace std      int main              string names        amz   Waq   Mon   Sam   Has   Shak   GBy       simple array         int length   sizeof names    sizeof names 0      give you size of array         int id          string append 7        as length is 7 just for sake of storing and printing output          for  size t i   0  i  lt  length  i                  id   rand     20000   2              append i    names i    to string id                     for  size t i   0  i  lt  length  i                  cout  lt  lt  append i   lt  lt  endl

User · Answer

There are a few options  and which one you want depends on the context   The simplest way is  std  cout  lt  lt  text  lt  lt  i    or if you want this on a single line  std  cout  lt  lt  text  lt  lt  i  lt  lt  endl    If you are writing a single threaded program and if you aren t calling this code a lot  where  a lot  is thousands of times per second  then you are done   If you are writing a multi threaded program and more than one thread is writing to cout  then this simple code can get you into trouble  Let s assume that the library that came with your compiler made cout thread safe enough than any single call to it won t be interrupted  Now let s say that one thread is using this code to write  Player 1  and another is writing  Player 2   If you are lucky you will get the following   Player 1 Player 2   If you are unlucky you might get something like the following  Player Player 2 1   The problem is that std  cout  lt  lt  text  lt  lt  i  lt  lt  endl  turns into 3 function calls  The code is equivalent to the following   std  cout  lt  lt  text  std  cout  lt  lt  i  std  cout  lt  lt  endl    If instead you used the C-style printf  and again your compiler provided a runtime library with reasonable thread safety  each function call is atomic  then the following code would work better   printf  Player  d n   i     Being able to do something in a single function call lets the io library provide synchronization under the covers  and now your whole line of text will be atomically written   For simple programs  std  cout is great  Throw in multithreading or other complications and the less stylish printf starts to look more attractive

User · Answer

cout  lt  lt  text  lt  lt  i

User · Answer

cout  lt  lt  text  lt  lt  i

User · Answer

cout  lt  lt  text  lt  lt  i    The  lt  lt  operator for ostream returns a reference to the ostream  so you can just keep chaining the  lt  lt  operations   That is  the above is basically the same as   cout  lt  lt  text  cout  lt  lt  i

User · Answer

printf  Player  d   i      Downvote my answer all you like  I still hate the C   I O operators     -P

User · Answer

cout  lt  lt   Player   lt  lt  i

User · Answer

cout  lt  lt  text  lt  lt       lt  lt  i  lt  lt  endl

User · Answer

cout  lt  lt  text  lt  lt  i    The  lt  lt  operator for ostream returns a reference to the ostream  so you can just keep chaining the  lt  lt  operations   That is  the above is basically the same as   cout  lt  lt  text  cout  lt  lt  i

User · Answer

cout  lt  lt  text  lt  lt       lt  lt  i  lt  lt  endl

User · Answer

cout  lt  lt  text  lt  lt  i    The  lt  lt  operator for ostream returns a reference to the ostream  so you can just keep chaining the  lt  lt  operations   That is  the above is basically the same as   cout  lt  lt  text  cout  lt  lt  i

User · Answer

If using Windows MFC  and need the string for more than immediate output try   int i   4  CString strOutput  strOutput Format  Player  d   i

User · Answer

There are a few options  and which one you want depends on the context   The simplest way is  std  cout  lt  lt  text  lt  lt  i    or if you want this on a single line  std  cout  lt  lt  text  lt  lt  i  lt  lt  endl    If you are writing a single threaded program and if you aren t calling this code a lot  where  a lot  is thousands of times per second  then you are done   If you are writing a multi threaded program and more than one thread is writing to cout  then this simple code can get you into trouble  Let s assume that the library that came with your compiler made cout thread safe enough than any single call to it won t be interrupted  Now let s say that one thread is using this code to write  Player 1  and another is writing  Player 2   If you are lucky you will get the following   Player 1 Player 2   If you are unlucky you might get something like the following  Player Player 2 1   The problem is that std  cout  lt  lt  text  lt  lt  i  lt  lt  endl  turns into 3 function calls  The code is equivalent to the following   std  cout  lt  lt  text  std  cout  lt  lt  i  std  cout  lt  lt  endl    If instead you used the C-style printf  and again your compiler provided a runtime library with reasonable thread safety  each function call is atomic  then the following code would work better   printf  Player  d n   i     Being able to do something in a single function call lets the io library provide synchronization under the covers  and now your whole line of text will be atomically written   For simple programs  std  cout is great  Throw in multithreading or other complications and the less stylish printf starts to look more attractive

User · Answer

cout  lt  lt   Player   lt  lt  i

User · Answer

Well  if you use cout you can just write the integer directly to it  as in  std  cout  lt  lt  text  lt  lt  i    The C   way of converting all kinds of objects to strings is through string streams  If you don t have one handy  just create one    include  lt sstream gt   std  ostringstream oss  oss  lt  lt  text  lt  lt  i  std  cout  lt  lt  oss str      Alternatively  you can just convert the integer and append it to the string   oss  lt  lt  i  text    oss str      Finally  the Boost libraries provide boost  lexical cast  which wraps around the stringstream conversion with a syntax like the built-in type casts    include  lt boost lexical cast hpp gt   text    boost  lexical cast lt std  string gt  i     This also works the other way around  i e  to parse strings

[c++] How do you append an int to a string in C++?

Examples related to c++

Examples related to int

Examples related to stdstring