Swapping pointers in C char int

Question

I have been struggling to understand the different behaviour when swapping pointers in C  If I want to swap two int pointers  then I can do  void intSwap  int  pa  int  pb       int temp    pa       pa    pb       pb   temp      However  if I want to swap two char pointers I need to do something like  void charSwap char   a  char   b       char  temp    a       a    b       b   temp      because if I do   void charSwap char  a  char  b       char temp    a       a    b       b   temp      the compiler complains about the expression  a    b as it cannot change the values  If I want to swap two strings  i e  char  s1   Hello   char  s2  Bye    how would one do it   Could you please give me a little bit of help  I would like to really learn how it works so I will not need to experience trial and error all the time until I get the right answer  I hope it s useful for many other people

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void intSwap  int  pa  int  pb       int temp    pa       pa    pb       pb   temp      You need to know the following -  int a   5     an integer  contains value int  p     an integer pointer  contains address p    amp a      amp a means address of a a    p      p means value stored in that address  here 5     void charSwap char  a  char  b       char temp    a       a    b       b   temp      So  when you swap like this  Only the value will be swapped  So  for a char  only their first char will swap   Now  if you understand char   string  clearly  then you should know that  you only need to exchange the pointer  It ll be easier to understand if you think it as an array instead of string   void stringSwap char   a  char   b       char  temp    a       a    b       b   temp      So  here you are passing double pointer because starting of an array itself is a pointer

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In C  a string  as you know  is a character pointer  char     If you want to swap two strings  you re swapping two char pointers  i e  just two addresses  In order to do any swap in a function  you need to give it the addresses of the two things you re swapping  So in the case of swapping two pointers  you need a pointer to a pointer  Much like to swap an int  you just need a pointer to an int   The reason your last code snippet doesn t work is because you re expecting it to swap two char pointers -- it s actually written to swap two characters   Edit  In your example above  you re trying to swap two int pointers incorrectly  as R  Martinho Fernandes points out  That will swap the two ints  if you had   int a  b  intSwap  amp a   amp b

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You need to understand the different between pass-by-reference and pass-by-value   Basically  C only support pass-by-value  So you can t reference a variable directly when pass it to a function  If you want to change the variable out a function  which the swap do  you need to use pass-by-reference  To implement pass-by-reference in C  need to use pointer  which can dereference to the value   The function   void intSwap int  a  int  b    It pass two pointers value to intSwap  and in the function  you swap the values which a b pointed  to  but not the pointer itself  That s why R  Martinho  amp  Dan Fego said it swap two integers  not pointers   For chars  I think you mean string  are more complicate  String in C is implement as a chars array  which referenced by a char   a pointer  as the string value  And if you want to pass a char  by pass-by-reference  you need to use the ponter of char   so you get char     Maybe the code below more clearly   typedef char  str  void strSwap str  a  str  b     The syntax swap int amp  a  int amp  b  is C    which mean pass-by-reference directly  Maybe some C compiler implement too   Hope I make it more clearly  not comfuse

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The first thing you need to understand is that when you pass something to a function  that something is copied to the function s arguments   Suppose you have the following   void swap1 int a  int b        int temp   a      a   b      b   temp      assert a    17       assert b    42          they re swapped     int x   42  int y   17  swap1 x  y   assert x    42   assert y    17      no  they re not swapped    The original variables will not be swapped  because their values are copied into the function s arguments  The function then proceeds to swap the values of those arguments  and then returns  The original values are not changed  because the function only swaps its own private copies   Now how do we work around this  The function needs a way to refer to the original variables  not copies of their values  How can we refer to other variables in C  Using pointers   If we pass pointers to our variables into the function  the function can swap the values in our variables  instead of its own argument copies   void swap2 int  a  int  b        int temp    a       a    b       b   temp      assert  a    17       assert  b    42          they re swapped     int x   42  int y   17  swap2  amp x   amp y      give the function pointers to our variables assert x    17   assert y    42      yes  they re swapped    Notice how inside the function we re not assigning to the pointers  but assigning to what they point to  And the pointers point to our variables x and y  The function is changing directly the values stored in our variables through the pointers we give it  And that s exactly what we needed   Now what happens if we have two pointer variables and want to swap the pointers themselves  as opposed to the values they point to   If we pass pointers  the pointers will simply be copied  not the values they point to  to the arguments   void swap3 int  a  int  b        int  temp   a      a   b      b   temp      assert  a    17       assert  b    42          they re swapped    void swap4 int  a  int  b        int temp    a       a    b       b   temp      assert  a    17       assert  b    42          they re swapped     int x   42  int y   17  int  xp    amp x  int  yp    amp y  swap3 xp  yp   assert xp     amp x   assert yp     amp y   assert x    42   assert y    17      Didn t swap anything  swap4 xp  yp   assert xp     amp x   assert yp     amp y   assert x    17   assert y    42      Swapped the stored values instead    The function swap3 only swaps its own private copies of our pointers that it gets in its arguments  It s the same issue we had with swap1  And swap4 is changing the values our variables point to  not the pointers  We re giving the function a means to refer to the variables x and y but we want them to refer to xp and yp   How do we do that  We pass it their addresses   void swap5 int   a  int   b        int  temp    a       a    b       b   temp      assert   a    17       assert   b    42          they re swapped      int x   42  int y   17  int  xp    amp x  int  yp    amp y  swap5  amp xp   amp yp   assert xp     amp y   assert yp     amp x   assert x    42   assert y    17      swapped only the pointers variables   This way it swaps our pointer variables  notice how xp now points to y  but not the values they point to  We gave it a way to refer to our pointer variables  so it can change them   By now it should be easy to understand how to swap two strings in the form of char  variables  The swap function needs to receive pointers to char    void swapStrings char   a  char   b       char  temp    a       a    b       b   temp      assert strcmp  a   world      0       assert strcmp  b   Hello      0      char  x    Hello   char  y    world   swapStrings  amp x   amp y   assert strcmp x   world      0   assert strcmp y   Hello      0

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If you have the luxury of working in C    use this   template lt typename T gt  void swapPrimitives T amp  a  T amp  b        T c   a      a   b      b   c      Granted  in the case of char   it would only swap the pointers themselves  not the data they point to  but in most cases  that is OK  right

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This example does not swap two int pointers  It swaps the value of the integers that pa and pb are pointing to  Here s an example of what s going on when you call this  void Swap1  int  pa  int  pb       int temp    pa       pa    pb       pb   temp     int main         int a   42      int b   17        int  pa    amp a      int  pb    amp b       printf  quot --------Swap1--------- n quot        printf  quot a    d n b    d n quot   a  b       swap1 pa  pb       printf  quot a    d n    d n quot   a  a       printf  quot pb address     p n quot   pa       printf  quot pa address     p n quot   pb      The output here is  a   42 b   17 pa address    0x7fffdf933228 pb address    0x7fffdf93322c --------Swap--------- pa   17 pb   42 a   17 b   42 pa address    0x7fffdf933228 pb address    0x7fffdf93322c  Note that the values swapped  but the pointer s addresses did not swap  In order to swap addresses we need to do this  void swap2  int   pa  int   pb       int temp    pa       pa    pb       pb   temp     and in main call the function like swap2  amp pa   amp pb   Now the addresses are swapped  as well as the values for the pointers  a and b have the same values that the are initialized with The integers a and b did not swap because it swap2 swaps the addresses being being pointed to by the pointers   a   42 b   17 pa address    0x7fffddaa9c98 pb address    0x7fffddaa9c9c --------Swap--------- pa   17 pb   42 a   42 b   17 pa address    0x7fffddaa9c9c pb address    0x7fffddaa9c98  Since Strings in C are char pointers  and you want to swap Strings  you are really swapping a char pointer  As in the examples with an int  you need a double pointer to swap addresses  The values of integers can be swapped even if the address isn t  but Strings are by definition a character pointer  You could swap one char with single pointers as the parameter  but a character pointer needs to be a double pointer in order to swap the strings

[c] Swapping pointers in C (char, int)

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