Changing the contents of a pointer variable

Question

In this code:

#include <iostream>
int main()
{
const char* name = "Abc";
std::cout<<*name<<std::endl;
return 0;
}

How can I change the contents of the pointer variable, not to what it is pointing at?

And, why am I just getting A as output from this program?

Thanks.

Answer 1

You get a char because you are de-referencing a char* pointer.

You could change the contents by doing:

*a = 'B'; // your char* would contain "Bbc"
a[0] = 'B'; // your char* would contain "Bbc"
a[1] = 'B'; // your char* would contain "ABc"
a[2] = 'B'; // your char* would contain "AbB"

But, changing the contents of a string literal has an undefined behavior in C. So you should not do that. Instead, you need to populate your char* dynamically. With something like this:

char *a;
a = malloc(sizeof(char)*100); // a string with a maximum of 99 chars
strcpy(a, "Abc");
// now you can safely change char* contents:
a[0] = '1';
a[1] = '1';
a[2] = '1';
printf("%s", a); // will print 111

String manipulation is not that easy in languages like C. Since by the syntax you are using, you seem to be compiling your code with a C++ compiler, then you might give string class a try instead of using pointers to char to handle your strings.