Question

I have here char text[60];

Then I do in an if:

if(number == 2)
  text = "awesome";
else
  text = "you fail";

and it always said expression must be a modifiable L-value.

Answer 1

lvalue means "left value" — it should be assignable. You cannot change the value of text since it is an array, not a pointer.

Either declare it as char pointer (in this case it’s better to declare it as const char*):

const char *text;
if(number == 2) 
    text = "awesome"; 
else 
    text = "you fail";

Or use strcpy:

char text[60];
if(number == 2) 
    strcpy(text, "awesome"); 
else 
    strcpy(text, "you fail");