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Why passing &error instead of error in Cocoa programming?

I have a question for which I cannot seem to find an answer…

I am using the SBJsonParser and there is a line of code I find puzzling:

NSError *error;
self.jsonData = [jsonParser objectWithString:responseString error:&error];

What is the & in front of the error parameter? (&error)?

Answer 1

In Objective-C, just like in C, & is the “address-of operator” and it returns the address of its argument. To find out more about it, I recommend you read this short chapter from The C Book.

Here’s an example of how the operator is used, to get a better idea:

#include <stdio.h>

// define a function that takes a pointer to an integer as argument
void change_value_of_int(int* int_to_change) {
    // change the value to which the argument points
    *int_to_change = 5;
}

int main() {
    // create a stack variable
    int test_int = 10;

    // pass the address of test_int to the function defined earlier
    change_value_of_int(&test_int);

    // now the value of test_int is 5
    printf("%d\n", test_int);

    return 0;
}

Note that the change_value_of_int() function expects the first parameter to be a pointer to an int, not an int, so you can’t call it with change_value_of_int(test_int). You must send it the address of the test_int variable, not the variable itself (because if you send a copy of the variable, it can’t change it).

Same thing with the NSError* example. jsonParser expects the address of a NSError*, not a NSError*, therefore the method is defined as:

- (id)objectWithString:(NSString*)jsonrep error:(NSError**)error;

Take a look at the header file and at the implementation to see how it is used. The value of your error (*error = the value of the thing pointed to by the error argument) becomes the return value of [errorTrace lastObject].