CharacterPointer in C++ - c++

Question

I have a piece of code as follows:
char* foo(char* str1)
{
str1 = "Some other text";
cout << "String Inside Function::" << str1 << endl;
return str1;
}
int main()
{
char* str = "This is a string";
cout << "String Before Function call::" << str << endl;
foo(str);
cout<<"String After Function call::"<<str<<endl;
return EXIT_SUCCESS;
}
But the cout after my function call gives me "This is a string" even though I have changed it in my foo function. I'm confused here although I know it has got something to do with me not passing the correct address.

Answer 1

You are changing the pointers value and not what it points to.

Answer 2

// Adding & (aka pass by reference) after the char* you can modify the pointer that you passed into foo function
// but you must understand that it was another place in memory!!!
void foo(char*& str1)
{
// This string saved in global section and here you
// changed not the text in the str1 but the pointer itself
str1 = "Some other text";
cout << "String Inside Function::" << str1 << endl;
}
// Another way to change pointer itself is pass a pointer to the pointer
void foo_v2(char** str1)
{
// You should dereference pointer to pointer (* before str1)
*str1 = "Some other text";
}
// In this case you change the content at the pointer str1
// It very dangerous, you can replace content not only the memory under str1
// but even return address if string placed in stack memory.
// Such terrible things occurred if your string that copied into str1
// occupied more bytes then it can be contained in str1
// More safe way is using for example std::string
void foo2(char* str1)
{
char *s = "Some other text";
strcpy(str1, s, strlen(s));
}
int main()
{
char* str = "This is a string";
cout << "String Before Function call::" << str << endl;
foo(str);
cout<<"String After Function call::"<<str<<endl;
return EXIT_SUCCESS;
}

Answer 3

When you pass a pointer by copy, the function you call gets a copy of that pointer, just as passing by copy of any other type means you can't modify the original, passing by pointer works in the same way.
void f(char* copy)
{
// the copy here is modified to point to test
copy = "Test";
}
void g(char* copy)
{
g[0] = 'C';
}
int main()
{
char* p = nullptr;
// f gets a copy of p, p cannot be modified by f
f(p);
char p2[5] = "copy";
// g gets a copy of p2, the value of p2 (the pointer) cannot be modified by g
// however, g can modify what p2 points to
g(p2);
// prints "Copy" (not "copy")
cout << p2;
}
If you want to change what a pointer "points to" via a function, you have to do that in the same way that you would change any other type passed to a function. Either through a pointer (double pointer in this case) or reference (preferred).
void make_c(char** c)
{
*c = new char[10];
}
void make_c(char*& c)
{
c = new char[10];
}
int main()
{
char* t = nullptr;
// via pointer (double pointer in this case)
make_c(&t);
delete [] t;
// via reference
make_c(t);
delete [] t;
}