Question

I’ve got the following code:

#include <iostream> using namespace std; int main() {     char* a = 'foo';     char* b = 'bar';     a = b;     cout << a << ', ' << b << endl;     return 0; } 

This compiles and works, ie. prints bar, bar. Now I would like to demonstrate that what goes on here is not copying a string. I would like to change b and show that a also changes. I came up with this simple code:

#include <iostream> using namespace std; int main() {     char* a = 'foo';     char* b = 'bar';     a = b;     b[1] = 'u'; // ← just this line added     cout << a << ', ' << b << endl;     return 0; } 

…but it segfaults. Why? The interesting thing is that the following modification runs just fine:

#include <iostream> using namespace std; int main() {     char* a = 'foo';     char b[] = 'bar'; // ← declaration changed here     a = b;     b[1] = 'u';     cout << a << ', ' << b << endl;     return 0; } 

Why doesn’t it segfault like the previous one? I guess I am missing some important difference between the pointer-style and the array-style string initialization.

Answer 1

You cannot change string constants, which is what you get when you use the pointer-to-literal syntax as in the first code samples.

See also this question: Is a string literal in c++ created in static memory?.