Why is this simple program not working?

[karan173]

 
#include<stdio.h>
 
int main()
{
    char *s="hello";
    *(s+2)='a';
    printf("%s",s);
    return 0;
}
 

it gives a runtime error, however if i declare s as a character array like s[]="hello", it works, why?

[manontheedge]

the problem is in the line

char *s="hello";

when you create a dynamic array, as you have done with

char* s

you need to reserve space in memory for it. You do that like this ...

char* s = new char[sizeof("hello")];

then, you can't assign a string of characters ("hello") to a dynamic char array using the equals sign. You have to copy the characters like this

strcpy(s, "hello");

... should work aside from that

[weaknessforcats]

Also,

char *s="hello";

says s is the address of the h, which is const. It's const becuse "hello" is a const array since it is a literal.

When you try to change this literal at run time you crash trying to change a const value.

You may do as manontheedge recommends rvided you delete the array.

If you need a local variable cod it this way:

char s[] ="hello";

Now this says that s is a char array allocated on the stack and initialized to hello+\0. The array has 6 elements.