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const qualifier in structure

Posted on 2009-04-09
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Last Modified: 2012-05-06
I want to know why the const qualifier in front of the array at the start of this program, supposedly makes the program more reliable and readable. I think it's because the specifiers in the printf statements have an easier time accessing the contents of an array if the array itself is constant. Is that correct?
#include <stdio.h> #define LEN 20 

const char * msgs[5] =

{ 

     "    Thank you,", 

     "You prove that a ",

     "is a special guy. We must meet over a "

     "at"  

     "and laugh"

}
 

struct names {

char first[LEN];

char last[LEN];

};
 

struct guy {

      struct names handle

       char favfood[LEN];

      char job[LEN] 

      float income;

};
 

int main(void)

{

struct guy fellow = {

      {"Thomas", "Villard"},

     "grilled salmon",

      "life coach",

      58112.00

};
 
 
 

     printf("Dear %s, \n\n", fellow.handle.first); 

     printf("%s%s.\n", msgs[0], fellow.handle.first);

     printf("%s%s\n", msgs[1], fellow.job);

     printf("%s\n", msgs[2]);

     printf("%s%s%s", msgs[3], fellow.favfood, msgs[4]);

     if (fellow.income > 150000.0) 

         puts("!!"); 

      else if (fellow.income > 75,000)

         puts("!");

      else

               puts(".");

      printf("See you soon.\n");
 

return 0;

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Question by:prebek
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mrjoltcola earned 500 total points
Comment Utility
Nothing to do with printf().

const specifies that the data shall not be changed. It encourages against bad coding practices, because if the compiler is strict, it won't allow passing a const as a non-const argument, without the programmer explicitly casting.

It simply tells that the array is of pointers to strings char are essentially readonly. A "const" string or array cannot be assigned to without casting to hide the true type of pointer, and even then, could result in a code crash even if, due to attempt to write to read-only memory.

The difference is illustrated below. It is not the array itself that is const, it is an array of pointers TO const strings.
const char * msgs[2] = { "test", "test" };
 

int main() {

  msgs[0] = "ABC";   // legal, can point to a NEW const string

  msgs[0][1] = 'A';  // illegal, what msgs[0] points to is not writeable

}

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by:prebek
Comment Utility
thanks, i get it now
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