[Solved] C does not support passing a variable by reference. How to do it?

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You’re right, C does not support passing by reference (as it is defined by C++). However, C supports passing pointers.

Fundamentally, pointers are references. Pointers are variables which store the memory address at which a variable can be located. Thus, standard pointers are comparable C++ references.

So in your case, void Foo(char *k, struct_t* &Root) would be similar to void Foo(char *k, struct_t **Root). To access the Root structure within the Foo function, you could then say something like:

void Foo(char *k, struct_t **Root){
    // Retrieve a local copy of the 1st pointer level
    struct_t *ptrRoot = *Root;
    // Now we can access the variables like normal
    // Perhaps the root structure contains an integer variable:
    int intVariable = ptrRoot->SomeIntegerVariable;
    int modRootVariable = doSomeCalculation(intVariable);
    // Perhaps we want to reassign it then:
    ptrRoot->SomeIntegerVariable = modRootVariable;
}

Thus, just passing pointers is equivalent to passing a reference.

solved C does not support passing a variable by reference. How to do it?