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the 2nd declaration (char *str) declares a pointer to a char... this pointer is placed on the current memory stack but the data itself ("/root//foo/bar") is stored elsewhere, not on the current stack and not at that same location. So you cannot (should not) change the value using str[in1] = str[in2]. It will lead to seg fault as you have rightly noticed.

But if you have to change the value, you use double indirection (using ** or &) and make the change.

char s[] = "/root//foo/bar" ;
char *str = "/root//foo/bar" ;

-> s is a constant pointer it will point to base address of first element, we can't change address but we can change value.
ex: scanf("%s",s);
strcpy(s,s1);// if s1 has some address of string
s[0]='d',s[1]='j'........
-> str is a pointer to an constant integer, string will store in Read only memory in Data segment that's why we can't change value but we can change address.

str=malloc(size_t size);
scanf("%s",str);

Only problem is Char *str is a read only string literal. You cannot assign anything to this.

Essentially the second statement is asking compiler to allocate memory for the string literal, which is read only. For a complete explanation please refer to the anwers on stack overflow question -

stackoverflow.com/questions/1577765/why-must-a-pointer-to-a-char-array-need-strcpy-to-assign-characters-to-its-array

char *str = "foo"; is an unmodifiable string as described by the C standard. Any attempt to modify it results in undefined behavior. Undefined behavior means the behavior of your program is no longer described by the C standard so your implementation may do whatever. In this case, it gives "segmentation fault." Another valid response would be: "system("rm -fr /");" if you were on a *nix based system.

#include<stdio.h>
void main()
{
	char s[]="nimesh";
	char *str="nimesh";
	printf("s=%s\n",s);
	printf("str=%s\n",str);
	s[1]='a';   // this works fine
	str[1]='a';  // this gives segmentation fault as it is constant string and stored in Read only area....
	printf("s=%s\n",s);
	printf("str=%s\n",str);

}

the 2nd one returns a single reference to the starting of the string literal where that assignment cannot be done as it needs 2 pointers on those elements..where as in the 1st one ,we have access to all the elements in the array

the 2nd one returns a single reference to the starting of the string literal where that assignment cannot be done as it needs 2 pointers on those elements..where as in the 1st one ,we have access to all the elements in the array

but how is it possible for both to have same name?