I have here char text[60];
Then I do in an if:
if(number == 2)
text = \"awesome\";
else
text = \"you fail\";
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");