CODEWITHSUNDEEP
c++parametersreference
pythonscripter
pythonscripter

Reputation:

C++ Parameter Reference

void (int a[]) { a[5] = 3; // this is wrong? }

Can I do this so that the array that is passed in is modified?

Sorry for deleting, a bit new here...

I have another question which might answer my question:

If I have

void Test(int a) {
}

void Best(int &a) {
}

are these two statements equivalent?

Test(a);
Best(&a);

Upvotes: 1

Views: 740

Answers (6)

James Curran
James Curran

Reputation: 103485

void Test(int a[]) 
{
    a[5] = 3;
}

just alternate syntax for:

void Test(int* a) 
{
    *(a+5) = 3;
}

No array is passed, just a pointer. The original array is modified.

As for your second revision, given:

void Test(int a) 
{
}

void Best(int &a) 
{
}

then 

Test(aa);      // Passes aa by value.  Changes to a in Test() do not effect aa
Best(aa);      // Passes aa by reference; Changes to a DO effect aa
Best(&aa);     // Is a syntax error: Passing a pointer instead of an int.

Upvotes: 13

Loki Astari
Loki Astari

Reputation: 264351

Your original Function should work.
If you give it a name:

#include <iostream>

// Arrays always de-generate to pointers.
void plop(int a[])  // Make sure this function has a name.
{
    a[5]    = 3;
}

int main()
{
    int test[]  = { 1,1,1,1,1,1,1,1};

    plop(test);

    std::cout << test[5] << std::endl;
}

This is because arrays always de-generate into pointers when passed as an argument to a function. So this should always work as expected. Assuming you don't index beyond the end of the array. Inside plop there is no way to determine the size of the array passed.

Upvotes: 0

Jason Baker
Jason Baker

Reputation: 198537

The primary motivator for passing arrays by reference is to prevent stack overflows and needless copying of large objects. For example, imagine if I had a function like this:

void foo(int x[500000000000]);

The stack would probably overflow the first time you called the function if all arrays were passed by value (but of course this is an obvious exaggeration).

This will become useful when using object-oriented methods. Suppose instead of an array, you had this:

void foo(SomeClass x);

where SomeClass is a class with 500000000000 data members. If you called a method like this, the compiler would copy x bit by bit, which would be a very long process to say the least. The same concept as you use in arrays still applies, but you have to specify that this is to be used by reference manually:

void foo(SomeClass &x);

(and don't go trying to create a 500000000000 element array to begin with unless you have a 64 bit machine and lots of RAM)

Upvotes: -1

Johannes Schaub - litb
Johannes Schaub - litb

Reputation: 506857

Read the answer given here about the difference of int[] and int* in a parameter list: Difference between char* and char[] . I've really put so much love into that answer! :)

Regarding your question about your Test and Best functions, James Curran provided an excellent answer.

Upvotes: 0

Drew Hall
Drew Hall

Reputation: 29047

Nope.

Your options for altering a value from outside the function are call by reference f(int& a), call by pointer f(int* a), and using a global (shudder...) variable.

Upvotes: 0

Uri
Uri

Reputation: 89729

If you get the variable not by reference and not by pointer, it means that the function is essentially isolated, getting an ad-hoc copy of a. No matter what you do (without trying to hack the stack or things like that) you wouldn't have access to that value in the calling context.

If you know something about the calling context, you may be able to do things based on some anticipation of stack contents, but it's generally a bad idea.

If your method takes a[] which is essentially a*, then yes, you can alter the contents of the cell that a points to, but you won't be able to alter a (the pointer) itself to point at something else.

Upvotes: 2

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