Question

I am working on Microsoft Visual Studio environment. I came across a strange behavior

 char *src ="123";
 char *des ="abc";
 printf("\nThe src string is %c", src[0]);
 printf("\tThe dest string is %c",dest[0]);

 des[0] = src[0];

 printf("\nThe src string is %c", src[0]);
 printf("\tThe dest string is %c",dest[0]);

The result is:

1 a
1 a   

That means the des[0] is not being initialized. As src is pointing to the first element of the string. I guess by rules this should work.

Answer 1

Since src and des are initialized with string literals, their type should actually be const char *, not char *; like this:

const char * src ="123";
const char * des ="abc";

There was never memory allocated for either of them, they just point to the predefined constants. Therefore, the statement des[0] = src[0] is undefined behavior; you’re trying to change a constant there!

Any decent compiler should actually warn you about the implicit conversion from const char * to char *…

If using C++, consider using std::string instead of char *, and std::cout instead of printf.