A weird declaration: int *a="any_string"

[shadyabhi]

#include<stdio.h>
int main(void)
{
	int *p="Hello_i_am_abhijeet";	//Whats this actually doing???
	char q[]="Hello_i_am_abhijeet";	//This is OK
	
	printf("%c %c\n",p[2],q[2]);	      //Here p[2] shows "a" that means 
						      //due to p[2], pointer moves 8 bytes 
	return 0;
}

Terminal Output:

codefire@codefi re-desktop:~$ gcc test.c
test.c: In function ‘main’:
test.c:4: warning: initialization from incompatible pointer type
codefire@codefi re-desktop:~$ ./a.out
a l
codefire@codefi re-desktop:~$


I want to know as to what actually happens in line 4 of program.
I know line 4 may look weird.
But, i was curious to know. And hows is the output coming, i mean whats theory behind that.

Also, it would be great if you pls help me understanding the difference between "char *a" and "char a[]". I tried searching on net but couldnt get satisfied with what i understood.
Your reply will be of great help to me.

[weaknessforcats]

Yes, it's mighty weird. Like it doesn't compile. "Hello_i_am_abh ijeet" is a literal string which is a char array. The address of the H is a char*. You cannot assign a char* to an int*.

If your compiler accepts this, then get a proper compiler.

Next:

char* c = "Hello_i_am_abhijeet";
char  d[] = "Hello_i_am_abhijeet";

In this example c is a pointer to a char. That char is the H of Hello. The literal itself is a char array but is constant so you cannot chnage the letters in the array.

On the other hand, d is an array of char but the number of elements is not provided. In this case the compiler counts the letters in "Hello_i_am_abh ijeet" and adds 1 for a null terminator. Then it creates the d array for that number oc characters and copies "Hello_i_am_abh ijeet" to the d array. This time d is not constant so you can change the letters in the array. This only works when d is created. Later on if you try:

d  = "New letters";  //ERROR
d[] = "New letters";  //ERROR

you get errors. To change d you would copy in new values:

strcpy(d, "New letters");

strcpy does not check that the new letters will fit in the d array. It's up to you to manage this.

[shadyabhi]

@weaknessforcat s

I used GCC as it can be understood through the terminal output i posted above..

But, still i didnt get the answer of what I actually asked.
Can anybody answer? I am really curious about this peculiarity.

[shadyabhi]

@weaknessforcat s

I used GCC as it can be understood through the terminal output i posted above..

But, still i didnt get the answer of what I actually asked.
Can anybody answer? I am really curious about this peculiarity.

[donbock]

I want to know as to what actually happens in line 4 of program.

Nothing happens in line 4 -- it doesn't compile.

[donbock]

Also, it would be great if you pls help me understanding the difference between "char *a" and "char a[]". I tried searching on net but couldnt get satisfied with what i understood.
Your reply will be of great help to me.

char *a = "This is a string";
char b[] = "So is this";

In the first case, the string (characters + null termination) is placed in an unnamed location and the address of that location is stored in variable a. sizeof(a) = the size of a pointer (typically 4 bytes).

In the second case, the string (characters + null termination) is placed in a buffer starting at location b. sizeof(b) = 11.

[shadyabhi]

thanx for reply ...
But, i tell you dude, that program(the exact code which i posted) compiles gcc 4.2.4...
Which compiler are you using??

[newb16]

>But, i was curious to know. And hows is the output coming, i mean whats theory behind that.

Assuming that program executes on little-endian machine, printf("%c", p[2]) prints ascii value of the lowest byte of p[2], and the address of this byte is the same as address of p[2]. As sizeof(int) is most likely 4, it's the byte with an offset of 8 from the lowest byte of p[0], and lowest byte of p[0] is q[0].

[weaknessforcats]

Then your gcc is using non-standard syntax rules.

You might try the -pedantic switch to disable non-standard features.

However, if you have the address of a char stored in an int* and you increment the int*, the compiler will advance sizeof(int) rather than sizeof(char) to totally screw up your memory.

So, if this is compiling, it is out of gross negligence to the pointer arithmetic rules.

[donbock]

Terminal Output:
codefire@codefi re-desktop:~$ gcc test.c
test.c: In function ‘main’:
test.c:4: warning: initialization from incompatible pointer type
codefire@codefi re-desktop:~$ ./a.out
a l
codefire@codefi re-desktop:~$

thanx for reply ...
But, i tell you dude, that program(the exact code which i posted) compiles gcc 4.2.4...
Which compiler are you using??

gcc 4.2.4 reports warning: initialization from incompatible pointer type. It isn't any happier about line 4 than I am. True, I overstated matters when I said "it doesn't compile". You get a compiler warning, not a compiler error; but that isn't good news.