Changing an argument from inside a function

Question

Consider functions A, B and C.

void A(){
  int arg;
  B(arg);
}

void B(int& arg){
  C(arg)
}    

void C(int& arg){
  arg = 10;
}

I want value of my argument to be set by function C. This code gives an error. The order of the function calls has to be A calls B, which calls C. How to do this?

Have you *declared* (i.e. made function prototypes of) the functions before you try to call them? It doesn't matter where they are defined (i.e. where the actual function implementation is) but they have to be declared before they are called. — Some programmer dude, Jan 14 '14 at 10:50

ereOn · Answer 1 · 2014-01-14T12:50:21.550

This should work, but you would have to write it that way (reorder the functions definitions):

void C(int& arg){
  arg = 10;
}

void B(int& arg){
  C(arg);
}    

void A(){
  int arg;
  B(arg);
}

Or to forward declare the functions before:

void B(int&);
void C(int&);

void A(){
  int arg;
  B(arg);
}

void B(int& arg){
  C(arg);
}    

void C(int& arg){
  arg = 10;
}

This way, A() knows of B() which knows of C().

score 0 · Answer 2 · answered Jan 14 '14 at 10:49

0

Reorder the functions in the source: C B A not A B C.

answered Jan 14 '14 at 10:49

Wojtek Surowka

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score 0 · Answer 3 · answered Jan 14 '14 at 10:52

Add prototype before your A() function. In your code, during compilation of A(), compiler have no idea about B(); So by function prototype you are telling compiler about it. And similarly for C()

void B(int&); 
void C(int&);

void A(){
    int arg;
    B(arg);
}

void B(int& arg){
    C(arg);
}    

void C(int& arg){
    arg = 10;
}

int main() {
    int a=20;
    A();
    return 0;
}

Changing an argument from inside a function

3 Answers3