C Programming - Strings - Discussion

Option 2 is wrong. Because p is a no constant pointer but it is pointing to the constant string (that string will be stored in the data segment in read only memory location).

I think option C is correct. Because in char *p="Sanjay". Pointer p is in stack but that pointer pointing string 'Sanjay" is store in read-only text section. So it is a const string. That too char a[]="Sanjay" in that statement a is not a const pointer. So I think option C is correct.

@All.

The program is to swap the first alphabet of the two strings. Can anyone tell the mistake?


#include<stdio.h>
int main(int argc, char* argv[])
{
char *c;
char *c1;
char *temp;
c=argv[1];
c1=argv[2];
temp[0]=c[0];
c[0]=c1[0];
c1[0]=temp[0];
printf("%s",c);
printf("%s",c1);
return 0;
}

@GAURAV.

In *str also the NULL char is added.

For e.g: We undergo to find the length of string

char *str="BALAJI";
int length;
while(*str!='\0')
{
length++;
str++;
}
return(length);

In the above program, it clearly shows that only after detecting the '\0' on *str, the while() condition gets terminates, Hence the NULL char will be default. Don't confuse with Sizeof(). Statement is in both cases the '\0' will be ADDED at the end of the string "Sanjay" is TRUE.

I'm not understanding any answer please explain clearly.

I'm not understsanding second option please explain.

Depending on compiler first option(2) may or may not be correct.

In turbo C it stores string pointed by pointer during initialization in data section whereas in gcc its in code section which is read only memory!

"a" is not a pointer. It is declared as char.

What is the meaning of second statement given in the question?

@Ronil.

I agree with your comment. NULL character will not added at the end of second array string. Because array of string by default added NULL character at the end of string.

For eg:

*str1="GAURAV";
*str2[]="GAURAV";

sizeof (str1) = 6;
sizeof (str2) = 7;