I have following piece of code where I am getting an error of
error: expected expression before const (at line 15)
12 : int
13 : function1(const char *arg1, const char **arg2)
14 : {
15 : int i = function2(const char *arg1, const char **arg2);
16 : }
18 : int
20 : function2(const char *arg1, const char **arg2)
21 : {
22 : }
what does that exactly mean?
Thanks
Remove int i = function2(const char *arg1, const char **arg2);
With int i = function2(arg1,arg2);
Your variables are already defined. when calling to function in C you shouldn't say parameters type , you have to pass parameters themselves.
Line 15 is a mistake. In C, variables are identified by a single token called an identifier.
In this case, the names of the variables are arg1 and arg2. You use those tokens when you are using the variables; you don't repeat all of the type information associated with the variable.
So the line should be:
int i = function2(arg1, arg2);
You already have arg1 and arg2 being passed in to your function1, so you just need
int i = function2(arg1, arg2);
int i = function2(const char *arg1, const char **arg2);
You are trying to call function2, but you have written the call like a function declaration. You don't specify the argument types when calling the function.
Remove the * and try? It seems like you're trying to pass arg1 and arg2 into function 2, but I don't think they can be referenced that way inside the function, because they weren't defined there. I believe it should be just
function2(arg1,arg2)
error: expected expression before const (at line 15)
Q What does that exactly mean?
A It means that the first thing after function2( is expected to be an expression. const is not an expression. arg1 will be an expression.
int i = function2(arg1,arg2);