CODEWITHSUNDEEP
cpointersmemory-managementheap-memory
Usama
Usama

Reputation: 662

Why this program works?

I am a noob trying to improve at pointers in C. I have declared a ptr to ptr variable strarray, dynamically allocated the memory without even typecasting to (char *) and assigned a string to ptr to ptr variable instead of pointer, but programs gives output "ddddd" how does it works

//array of string 
#include<stdio.h>
#include<stdlib.h>
void main(){
    char **strarray;
    int arraylengh=10;
    strarray = malloc(sizeof(char *) * arraylengh);
    strarray[4]="ddddd";
    printf("%s ", strarray[4]);

}

strarray is a pointer to pointer variable and I am assigning string directly to it strarray[4]="ddddd"; strarray[4] should be assigned to pointer pointing to a string right? and i can see ddddd as the output

Upvotes: 1

Views: 150

Answers (3)

programmer
programmer

Reputation: 679

strarray = malloc(sizeof(char *) * arraylengh);

Here you allocate memory for a array of char * pointers.

The visualization below shows how this looks in memory:

strarray -> ptr1 | ptr2 | ptr3 | ptr4...

strarray is a pointer to pointer variable and I am assigning string directly to it strarray[4]="ddddd"; strarray[4] should be assigned to pointer pointing to a string right? and i can see ddddd as the output

Assigning a string to strarray[4] is perfectly valid. Since the array is made up of pointers, you are free too assign any pointer from strarray[0] to strarray[9] with a string.

On the following line:

strarray[4]="ddddd

You assign one of these pointers with a string. This is essentially the same as assigning a char * pointer with a string, for example:

char *a_string = "ddddd";

But the only difference is that strarray[4] is a part of the array of pointers you allocated.

The reason this works is because the string "ddddd" is a pointer in your code.

In C strings are arrays of characters which are NULL terminated. So when we reference the array of characters that makes up the string "ddddd", it will implicitly decay to a pointer to the first element of the array.

Upvotes: 2

Steve Summit
Steve Summit

Reputation: 48010

I find it easiest to understand these situations with pictures. By the time your program is finished, it has set up the pointer-to-pointer variable strarray so that it looks like this:

          +---------+
strarray: |    *    |
          +----|----+
               |
               |
               v
          +---------+
          |  ?????  |
          +---------+
          |  ?????  |
          +---------+
          |  ?????  |
          +---------+
          |  ?????  |
          +---------+        +---+---+---+---+---+----+
          |    *-----------> | d | d | d | d | d | \0 |
          +---------+        +---+---+---+---+---+----+
          |  ?????  |
          +---------+
          |  ?????  |
          +---------+
          |  ?????  |
          +---------+
          |  ?????  |
          +---------+
          |  ?????  |
          +---------+

When you called malloc, you got an unnamed block of memory suitable for containing 10 pointers. You assign strarray to point to that block of memory, which acts so much like an array it might as well be one. You set the fifth element of that "array" to point to the string "ddddd". The other 9 pointers remain indeterminate, indicated here by ?????.

The string "ddddd" is automatically allocated for you by the compiler, and you successfully manipulate a pointer to that string when you say

strarray[4] = "ddddd";

Upvotes: 3

Schwern
Schwern

Reputation: 165218

You're not copying "ddddd" to strarray[4], you are copying a pointer to "ddddd".

All strings in C are pointers, including "ddddd" (technically an array which "degrades" to a pointer). strarray[4]="ddddd" is not copying "ddddd", it is assigning a pointer.

char *str = "ddddd";
strarray[4] = str;

Same thing.

We can print them as pointers to see they point to the same thing.

printf("%p %p\n", strarray[4], str);

Upvotes: 2

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