C Pointers – Understanding 2D Arrays and Pointers

c++pointers

Hey All, I am a pointer newbie and in the following code, I am trying to store the values of a 2 D array in a structure and then print them. However, I get a compilation error at the line: fd->mychar[i] = newptr[i]; I get that while char * str is same as str[], char ** str isnt the same as str[][], but I cant find a solution to make the following work.

typedef struct mystruct{
  char mychar [20][20];
}mystruct_t;

void printvalues ( char ** newptr){
  int i;
  mystruct_t * fd;
  for (i=0;i<3;i++){
    fd->mychar[i] = newptr[i];
    printf("My value is %s and in struct %s\n", newptr[i], fd->mychar[i]);
  }
}
int main (int argc, char **argv){
 int i;
 char * abc[5] = {"123", "456", "789"};

 for (i=0;i<3;i++){
  printf("My value is %s\n", abc[i]);
 }
 printvalues(abc);
}

Best Answer

Most of the issue was your use of an unallocated structure. You used a pointer to mystruct_t but never allocated it. The following runs for me:

#include <stdio.h>

typedef struct mystruct
{
   char*  mychar [20];
} mystruct_t;

void printvalues( char** newptr )
{
   int i;
   // note: you used an unallocated pointer in your original code
   mystruct_t fd;
   for ( i = 0; i < 3; i++ )
   {
      fd.mychar[i] = newptr[i];
      printf( "My value is %s and in struct %s\n", newptr[i], fd.mychar[i] );
   }
}

int main( int argc, char **argv )
{
   int i;
   char * abc[5] = { "123", "456", "789" };

   for ( i = 0; i < 3; i++ )
   {
      printf( "My value is %s\n", abc[i] );
   }
   printvalues( abc );
}

Related Solutions

C Programming – Pointer to 2D Arrays in C

//defines an array of 280 pointers (1120 or 2240 bytes)
int  *pointer1 [280];

//defines a pointer (4 or 8 bytes depending on 32/64 bits platform)
int (*pointer2)[280];      //pointer to an array of 280 integers
int (*pointer3)[100][280]; //pointer to an 2D array of 100*280 integers

Using pointer2 or pointer3 produce the same binary except manipulations as ++pointer2 as pointed out by WhozCraig.

I recommend using typedef (producing same binary code as above pointer3)

typedef int myType[100][280];
myType *pointer3;

Note: Since C++11, you can also use keyword using instead of typedef

using myType = int[100][280];
myType *pointer3;

in your example:

myType *pointer;                // pointer creation
pointer = &tab1;                // assignation
(*pointer)[5][12] = 517;        // set (write)
int myint = (*pointer)[5][12];  // get (read)

Note: If the array tab1 is used within a function body => this array will be placed within the call stack memory. But the stack size is limited. Using arrays bigger than the free memory stack produces a stack overflow crash.

The full snippet is online-compilable at gcc.godbolt.org

int main()
{
    //defines an array of 280 pointers (1120 or 2240 bytes)
    int  *pointer1 [280];
    static_assert( sizeof(pointer1) == 2240, "" );

    //defines a pointer (4 or 8 bytes depending on 32/64 bits platform)
    int (*pointer2)[280];      //pointer to an array of 280 integers
    int (*pointer3)[100][280]; //pointer to an 2D array of 100*280 integers  
    static_assert( sizeof(pointer2) == 8, "" );
    static_assert( sizeof(pointer3) == 8, "" );

    // Use 'typedef' (or 'using' if you use a modern C++ compiler)
    typedef int myType[100][280];
    //using myType = int[100][280];

    int tab1[100][280];

    myType *pointer;                // pointer creation
    pointer = &tab1;                // assignation
    (*pointer)[5][12] = 517;        // set (write)
    int myint = (*pointer)[5][12];  // get (read)

    return myint;
}

Best Answer

Related Solutions

C Programming – Pointer to 2D Arrays in C

Related Question