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Reference variables as parameters: pass-by-reference

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• Passing parameters by reference
  • Recall that

    C passes a variable by value        This was discussed here: click here

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  • Fact:

    The reference operator & in C, enable the C programmer to pass a parameter by reference !!!

    Example:

    int main(int argc, char *argv[]) { int x1 = 4, x2 = 88; func( &x1 ); // Pass address of x1 to function func( &x2 ); // Pass address of x2 to function }

    Explanation:

    From this webpage on assembler programming ( click here), we saw that in pass-by-reference, the compiler will: Pass the address (= reference) of a variable to the function          Because: &x = the address of variable x &y = the address of variable y we have passed the variables x and y by reference in the above example code !!!

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• Reference variables as parameters of a function
  • In this example, we passed the address of the variable int a to the function func( ):

    func( &x );

    Question:

    How do we define this function ? Specifically: What is the data type of the parameter ???

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  • The answer can be found from the previous discussion where we saw that you can assign the address of a variable to a reference variable of the same data type:

    int a = 7; // an int typed variable // &a = the address of a int typed variable int *p; // p can hold an address of an int typed variable // Therefore: p can store &a !!! p = &a; // Store the address of the int type variable a in p

    So we must use:

    int *varName

    as parameter type !!!

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  • Example: this function func( ) will print the value of the parameter and the value in referenced variable

    #include <stdio.h> void func( int *p ) { printf( " p = %u\n", p ); // Print the parameter p // We SHOULD use %p (hex), but I like %u (dec) // I will get a warning and I can live with it printf( "*p = %d\n\n", *p ); // Print the variable pointed to by p } int main(int argc, char *argv[]) { int x1 = 4, x2 = 88; func( &x1 ); // Pass address of x1 to function func( &x2 ); // Pass address of x2 to function }

    The output of this program is:

    cs255-1@aruba (5572)> a.out p = 3731491632 *p = 4 p = 3731491636 *p = 88

    You can see that the function func( ) can access the original variables through the reference (= address) passed by &x and &y !!!

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• A working example of pass-by-reference
  • In the next example program, I will write a function tha increments the input parameter by 1

    I do so by:

    Pass the parameter variable by reference to the function func( ) Inside the function func( ), I use the de-reference operator * to access the original variable

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  • Here is the code for function func( ):

    #include <stdio.h> void func( int *p ) { // The expression *p will access the ORIGINAL variable !! *p = *p + 1; // Add 1 to the (original) variable } int main(int argc, char *argv[]) { int x1 = 4, x2 = 88; printf("x1 before = %d\n", x1); func( &x1 ); // Pass address of x1 to function printf("x1 after = %d\n", x1); printf("x2 before = %d\n", x2); func( &x2 ); // Pass address of x2 to function printf("x2 after = %d\n", x2); }

    Output of this program:

    cs255-1@aruba (5577)> a.out x1 before = 4 x1 after = 5 x2 before = 88 x2 after = 89

    So you can see that the parameter was passed by reference !!!

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  • Example Program: (Demo above code)                                                
    • Prog file: /home/cs255001/demo/C/set2/ref-param3.c

    How to run the program:

    To compile:   gcc ref-param3.c          To run: ./a.out

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• Summary conclussion
  • Summary:

    We saw that C only provide pass-by-value.... However: C also has 2 special (low level) operators: & that returns the address of a variable * that accesses a variable using a reference (= address) Through these low level operators: C can enable the C programmer to pass any variable by reference !!!