Question

Can someone explain why this simple ternary operation won’t even compile, in C?

void main(int argc, char *argv[]){
    int a = atoi(argv[1]);
    char foo[] = (a == 1) ? "bar1" : "bar2";
}

It seems to be a problem with strings in particular.

Answer 1

A string literal "bar", when used in an expression (in this case, inside the ternary operator), is a pointer to pre-allocated memory. You cannot initialize an array using a pointer to data, only using a literal ("..." or {...}).

Instead, you could assign it to a char *:

const char *foo = (a == 1) ? "bar1" : "bar2";

This will not make a copy of the literal but point to it, so you should not modify the elements of foo. If you need a copy, you can use memcpy, provided that you know how big to declare the array:

char foo[5];
memcpy(foo, (a == 1) ? "bar1" : "bar2", sizeof foo);

If you particularly want to be able to assign the contents, then there is a trick to doing that; it is possible to implicitly copy the contents of a struct using assignment (as well as returning it from a function, and so on), no matter what the struct contains, and you could put a character array in a struct:

typedef struct { 
  char contents[5];
} mystring;

mystring foo = (a == 1) ? (mystring){"bar1"} : (mystring){"bar2"};
// You can also use assignments.
foo = (mystring){"baz"};

Just like the second option, when you do this you have to pick a fixed size for the array in the struct declaration. If you have strings of unbounded length, then you must use pointers.