A small thought on the Basics Again

[karthick88]

I was thinking about the following code

char *argv[] = { "Karthick", "88"};


so whats happening here. ??

Now to get you all clear of what my question is let me give an example. when we execute int a[] = { 12,13 }, the following would happen


a = &int(12); // note that the int(12) is called the address of which is assigned to a
a+1 = &int(13) // now new memory is allocated to 13 and that address is stored in a+1.

So the same what ll happen when the code in the beginning of the post is executed??

[jsmith]

Using your terminology,

argv is the address of the pointer to "Karthick",
argv1 is the address of the pointer to "88".

Though this statement isn't true:

a+1 = &int(13) // now new memory is allocated to 13 and that address is stored in a+1.

the address of the integer 13 is not stored in a+1; it is a+1.

[kempofighter]

You created an array of two char* and each points to a string literal which cannot be modified. If you try to modify them you will cause undefined behavior within your program. To be honest I am not sure what your question is because the two arrays in question are totally different. One is an array of pointers. One is an array of integers.

In other words
*a == 12
*(a+1) = 13.

However,
*argv = address of first character of "Karthick"
*(argv+1) = address of first character of "88"

So in the case of argv it is an indirect access of each string through the pointer stored.

[karthick88]

Oh ok.. Thanks Smith :)