How do pointers make a 1D array effectively 2D?

list is an array of 3 char * types. It's a 1D array of char pointers type, where it's each pointer is pointing to a string literal.

The in-memory view of list would be something like this:

      list                       
        +----+    +-+-+-+-+--+
 list[0]|    |--->|p|u|n|e|
      list                       
+----+    +-+-+-+-+--+
list[0]|    |--->|p|u|n|e|\0|
|    |    +-+-+-+-+--+
+----+    +-+-+-+-+-+-+-+-+--+
list[1]|    |--->|b|a|n|g|l|o|r|e|\0|
|    |    +-+-+-+-+-+-+-+-+--+
+----+    +-+-+-+-+-+-+-+--+
list[2]|    |--->|c|h|e|n|n|a|i|\0|
|    |    +-+-+-+-+-+-+-+--+
+----+    
|
        |    |    +-+-+-+-+--+
        +----+    +-+-+-+-+-+-+-+-+--+
 list[1]|    |--->|b|a|n|g|l|o|r|e|
      list                       
+----+    +-+-+-+-+--+
list[0]|    |--->|p|u|n|e|\0|
|    |    +-+-+-+-+--+
+----+    +-+-+-+-+-+-+-+-+--+
list[1]|    |--->|b|a|n|g|l|o|r|e|\0|
|    |    +-+-+-+-+-+-+-+-+--+
+----+    +-+-+-+-+-+-+-+--+
list[2]|    |--->|c|h|e|n|n|a|i|\0|
|    |    +-+-+-+-+-+-+-+--+
+----+    
|
        |    |    +-+-+-+-+-+-+-+-+--+
        +----+    +-+-+-+-+-+-+-+--+
 list[2]|    |--->|c|h|e|n|n|a|i|
      list                       
+----+    +-+-+-+-+--+
list[0]|    |--->|p|u|n|e|\0|
|    |    +-+-+-+-+--+
+----+    +-+-+-+-+-+-+-+-+--+
list[1]|    |--->|b|a|n|g|l|o|r|e|\0|
|    |    +-+-+-+-+-+-+-+-+--+
+----+    +-+-+-+-+-+-+-+--+
list[2]|    |--->|c|h|e|n|n|a|i|\0|
|    |    +-+-+-+-+-+-+-+--+
+----+    
|
        |    |    +-+-+-+-+-+-+-+--+
        +----+    

list is not storing 2D information of string but it's each element (which is a pointer) pointing to a string¹⁾.

Below part included, just in case, if you still confused and thinking of list as 2D array or storing 2D information.

Declaration of list as 2D array:

char list[3][9] = {"pune","banglore","chennai"};

Now, the list is a 2D array.
You can think of it's in-memory view something like this:

 index   0 1 2 3 4 5 6 7 8 
  list  +-+-+-+-+-+-+-+-+-+
list[0] |p|u|n|e|0|0|0|0|0|
        +-+-+-+-+-+-+-+-+-+
list[1] |b|a|n|g|l|o|r|e|0|
        +-+-+-+-+-+-+-+-+-+
list[2] |c|h|e|n|n|a|i|0|0|
        +-+-+-+-+-+-+-+-+-+

Note that, in this case the string literals used to initialize array, the characters of string literals are copied to array memory. But in the former case (where list is an array of char *), the string literal converted to pointer to its first element¹⁾.

From C11 Standards#6.3.2.1p3 [emphasis added]

>3 Except when it is the operand of the sizeof operator, the _Alignof operator, or the unary & operator, or is a string literal used to initialize an array, an expression that has type ''array of type'' is converted to an expression with type ''pointer to type'' that points to the initial element of the array object and is not an lvalue. ....

It is a 1D array of pointers to char.
Pointers to char are the closest C comes to "strings", i.e. pointers to the first of a sequence of characters which have their end/length indicated with a 0-terminator. That happens to be the way things like "pune" are stored/represented.

That array, which has three entries, gets initialised with "strings".
It ends up containing pointers to character sequences which should be considered read-only, as you are not supposed to write to them.

In the shown code a final ; seems to be missing, but I think that is just your intentional choice of copy-paste content and you do not have any trouble of understanding there.

"pune", "banglore" and "chennai" are called string literals. String literal is C sting. C string is a char array containing all the characters plus terminating null character (which is usually \0 [and it has zero integer value]). String arrays cannot be modified by your program.

Arrays in C decay to pointers (when you use array in expression it decays to pointer to the first element of the array). It happens here.

char *list[3]={"pune","banglore","chennai"} defines 1D array of pointers to char and initializes it with references to first characters of the string literals. It is equivalent to.

const char arr1[] = "pune";
const char arr2[] = "banglore";
const char arr3[] = "chennai";
char* list[3]={arr1, arr2, arr3}

To be more cont correct (as you can't modify the string literals) it should be declared as:

const char *list[3]={"pune","banglore","chennai"};

> char* list[3]={"pune","banglore","chennai"}

When a string literal, like "pune", appears in source code, a static array of characters is created for it.¹ That array contains the characters of the string (“p”, “u”, “n”, and “e”) and a terminating null character (code 0).

char * list[3] declares an array of three elements. Each element is a char *. A char * is a pointer to a char.

={"pune","banglore","chennai"} says to initialize the array with the expressions "pune", "banglore", and "chennai". "pune" is used to initialize list[0], "banglore" is used to initialize list[1], and "chennai" is used to initialize list[2].

When an array is used in an expression other than as the operand of sizeof, as the operand of unary &, or to initialize an array, it is automatically converted to a pointer to its first element. Here, "pune" is being used to initialize list[0], which is a pointer (char *), not an array. So the array represented by "pune" is converted to a pointer to its first element, “p”, and then list[0] is initialized with that pointer.

So list[0] is set to point to the first element of "pune". Similarly, list[1] is set to point to the first element of "banglore", and list[2] is set to point to the first element of "chennai".

So list is not a two-dimensional array. It is a one-dimensional array of pointers to char. The char it points to are in memory arranged by the compiler. They are not necessarily in some orderly arrangement like an array.

Footnote

¹ During compiler optimization, the contents of this array might be incorporated into the program and the separate array might be eliminated, depending on circumstances.