C Programming - Strings - Discussion

lvalue means location value.
This question has two parts.

<1> The Erroneous One:

For example: int a = 7;

Here, a = lvalue because it is modifiable. We can update "a" according to our wish.

And anywhere a = 8; is acceptable.

Now suppose we write: 7 = 8;

Can that be done?
Never! Because 7 is not a variable/lvalue.

Lets come back to the question. Here, "str" means a pointer pointing to the base of the array "str".

So, It's a constant. How can you assign a string to a constant? So, we are getting an error.


<2> The Correct One:

After using strcpy, we actually copied everything of the second string to the first one. It's just an analogy of
x = y; (where x and y are two integers).

Here is the working process of the correct code.

>> First, we had "Nagpur".

>> Then, we just changed str[0] or "N" with "K".

>> It leaves "Kagpur".

>> We print that.

>> Next, we update "Kagpur" to "Kanpur".

>> Now we have to understand how strings get printed.

printf("%s, ", str) --> This signifies that everything from address str is printed until '\0' is encountered.

>> So printf("%s ", str+1) just increments the pointer to point to the next character i.e "a" and prints "anpur".

Hope I can make it clear..

char str[] = "Nagpur";

str = "Kanpur";

Here str is a constant pointer and it is pointing to a string "Nagpur" (that is an array of char type).

In the second line we are trying to override the array which is not possible. we can assign it character by character. There is a inbuilt function strcpy(str,"Kanpur") that can be used for direct copying(overriding).

#include<stdio.h>

int main()
{
char str[] = "Nagpur";
str[0]='K';
printf("%s, ", str);
*str = "Kanpur";
printf("%s", str+1);
return 0;
}

This Gives O/P - Kagpur, agpur.

So We cannot just assign a value directly to the base address, We have to dereference to do that.

"String" is not a data type in C (unlike python, Java etc.) so you cannot simply assign:

char name[20];
name = "market"; // wrong
name[0] = "cricket"; // wrong again

"cricket" is called a string literal whereas name is a char array, i.e {'c','r','i','c','k','e','t','\0'}. They are not automatically converted to each other - you have to write your own function to copy char-by-char or use the library function strcpy().

Then how is;
char name[20] = "This is me";
valid code? Well, this is just a C short-cut for writing
char name[20] = {'T','h','i','s',' ','i','s',' ','m','e','u','\0'};

It does not work anywhere else.

Please, anyone, explain clearly.

Guys, the answer D is correct but the official explanation is horribly wrong!

This line is problematic:

"str[0]='K';"
The reason is, a literal string like "Nagpur" is stored in static memory area, READ ONLY. You can not change the value.

There is no problem to assign to str a different literal string because it is a char pointer, that is what a point is born to do.

Try run the code yourself, comment all lines after "str[0]='K';" and see what you find out.

No, it won't produces the error, the string can be assigned directly using = during the initialization. So the answer will be C).

An, Array is internally considered as Pointer. And the name of an array is internally a base address.

Please not understanding. Can you please explain in detail?

str holds the address of first character in an array. Its a pointer.

If you give str[0+1]->means str[1]-> it will prints "a" alone.

But I have an doubt that how str becomes an pointer? Its just an character array.

How can you do like str+1? Only we can do like str[0+1]. Anyone clear my doubt.