How to increment a pointer address and pointer s value

Question

Let us assume   int  p  int a   100  p    amp a    What will the following code will do actually and how       p      p     p      p       p    p      p       p        p      p     I know  this is kind of messy in terms of coding  but I want to know what will actually happen when we code like this    Note   Lets assume that the address of a 5120300  it is stored in pointer p whose address is 3560200  Now  what will be the value of p  amp  a after the execution of each statement

User · Answer

The following is an instantiation of the various "just print it" suggestions. I found it instructive.

#include "stdio.h"

int main() {
    static int x = 5;
    static int *p = &x;
    printf("(int) p   => %d\n",(int) p);
    printf("(int) p++ => %d\n",(int) p++);
    x = 5; p = &x;
    printf("(int) ++p => %d\n",(int) ++p);
    x = 5; p = &x;
    printf("++*p      => %d\n",++*p);
    x = 5; p = &x;
    printf("++(*p)    => %d\n",++(*p));
    x = 5; p = &x;
    printf("++*(p)    => %d\n",++*(p));
    x = 5; p = &x;
    printf("*p++      => %d\n",*p++);
    x = 5; p = &x;
    printf("(*p)++    => %d\n",(*p)++);
    x = 5; p = &x;
    printf("*(p)++    => %d\n",*(p)++);
    x = 5; p = &x;
    printf("*++p      => %d\n",*++p);
    x = 5; p = &x;
    printf("*(++p)    => %d\n",*(++p));
    return 0;
}

It returns

(int) p   => 256688152
(int) p++ => 256688152
(int) ++p => 256688156
++*p      => 6
++(*p)    => 6
++*(p)    => 6
*p++      => 5
(*p)++    => 5
*(p)++    => 5
*++p      => 0
*(++p)    => 0

I cast the pointer addresses to ints so they could be easily compared.

I compiled it with GCC.

User · Answer

With regards to  How to increment a pointer address and pointer s value   I think that     p     is actually well defined and does what you re asking for  e g     include  lt stdio h gt   int main       int a   100    int  p    amp a    printf   p n   void  p         p       printf   p n   void  p     printf   d n  a     return 0      It s not modifying the same thing twice before a sequence point  I don t think it s good style though for most uses - it s a little too cryptic for my liking

User · Answer

Note          1  Both    and   have same precedence priority   so the associativity comes into picture          2  in this case Associativity is from   Right-Left            important table to remember in case of pointers and arrays            operators           precedence        associativity      1                          1               left-right     2        identifier        2               right-left     3    lt data type gt             3               ----------          let me give an example  this might help           char   str          str    char    malloc sizeof char   2      allocate mem for 2 char          str 0   char   malloc sizeof char  10      allocate mem for 10 char         str 1   char   malloc sizeof char  10      allocate mem for 10 char          strcpy str 0   abcd        assigning value         strcpy str 1   efgh        assigning value          while  str                        cout lt  lt  str lt  lt endl       printing the string              str                   incrementing the address pointer                                     check above about the prcedence and associativity                   free str 0            free str 1            free str

User · Answer

First  the    operator takes precedence over the   operator  and the    operators take precedence over everything else   Second  the   number operator is the same as the number   operator if you re not assigning them to anything  The difference is number   returns number and then increments number  and   number increments first and then returns it   Third  by increasing the value of a pointer  you re incrementing it by the sizeof its contents  that is you re incrementing it as if you were iterating in an array   So  to sum it all up   ptr          Pointer moves to the next int position  as if it was an array    ptr        Pointer moves to the next int position  as if it was an array     ptr       The value of ptr is incremented     ptr      The value of ptr is incremented     ptr      The value of ptr is incremented  ptr         Pointer moves to the next int position  as if it was an array   But returns the old content   ptr        The value of ptr is incremented   ptr        Pointer moves to the next int position  as if it was an array   But returns the old content    ptr       Pointer moves to the next int position  and then get s accessed  with your code  segfault     ptr      Pointer moves to the next int position  and then get s accessed  with your code  segfault   As there are a lot of cases in here  I might have made some mistake  please correct me if I m wrong   EDIT   So I was wrong  the precedence is a little more complicated than what I wrote  view it here  http   en cppreference com w cpp language operator precedence

User · Answer

checked the program and the results are as   p          use it then move to next int position   p        move to next int and then use it    p       increments the value by 1 then use it      p      increments the value by 1 then use it     p      increments the value by 1 then use it  p         use the value of p then moves to next position   p        use the value of p then increment the value   p        use the value of p then moves to next position    p       moves to the next int location then use that value     p      moves to next location then use that value

[c] How to increment a pointer address and pointer's value?

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