// 1) This function takes an array as an argumentvoid funcArray(char arg[5]);// 2) This one takes a pointervoid funcPointer(char *arg);// 3) And this on etakes a *reference* to an arrayvoid funcRefArray(char (&arg)[]);// 4) and finally... this one takes a reference to a pointervoid funcRefPointer(char *(&arg));Quote:Original post by swiftcoderQuote:Original post by rip-offQuote:Original post by swiftcoder
In this example, there is no need to pass the array by reference. Don't forget that arrays in C are really *just* pointers (extra syntax non-withstanding), so any function passed an array can modify the array's contents.
Arrays decay to pointers, arrays are not pointers.
nitpicker :)
If I declare 'char A[5]', A is a pointer. Albeit to a either the static data section or to a region on the stack, and immutable.
Quote:Quote:Quote:You only need to pass a reference to an array if you need the function to change the array pointer itself (i.e. pointing it at another buffer), and that doesn't work for static arrays.
If you pass a reference to an array, you can use sizeof on the array and not get sizeof(pointer).
But in fact that is just a syntactical issue, and you *have* created an unnecessary temporary pointer. And are you quite sure about that? I am not near a compiler to test that, but it seems to me you should get the same result whether you pass an array, or a reference to an array (note that I never mentioned using a pointer directly).
Quote:Original post by swiftcoder
If I declare 'char A[5]', A is a pointer. Albeit to a either the static data section or to a region on the stack, and immutable.
Quote:But in fact that is just a syntactical issue, and you *have* created an unnecessary temporary pointer. And are you quite sure about that? I am not near a compiler to test that, but it seems to me you should get the same result whether you pass an array, or a reference to an array (note that I never mentioned using a pointer directly).
#include <iostream>int byref(int (&a)[5]){ return sizeof(a);}int byval(int a[5]){ return sizeof(a);}int main(){ int a[5]; std::cout << byref(a) << std::endl << byval(a) << std::endl;}204Quote:Original post by swiftcoder
It looks as if several posters may have confused the terms 'array', 'pointer', and 'reference' in this context.
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I never mentioned the pointer variants, I was talking about 1) and 3), and pointing out that in either the fnction can modify the array contents.
void example( char arg[] );void foo() { char array[4] = {}; example(array); // array decays to a pointer char *pointer = array; example(pointer); // look we can pass a pointer directly // this wont // array++; }void example( char arg[] ) { // look what we can do now! arg++;}template< class T, int N >int array_size( T (&arg)[N] ) { return N;}void bar( int arg[] );void foo() { int array[5]; int x = array_size(array); // x == 5 bar(array); // wont work because...}void bar( int arg[] ) { int x = array_size(arg); // ...no function to match array_size( int *& )}Quote:Original post by ToohrVyk
The sequence of operations performed for accessing A[3] is:
- Get the address of A, add 3, dereference (if A is an array).
- Get the address of A, dereference, add 3, dereference (if A is a pointer).
int foo(int A[], int n) { return A[n];}int bar(int * A, int n) { return A[n];} mov eax, DWORD PTR _n$[esp-4] mov ecx, DWORD PTR _A$[esp-4] mov eax, DWORD PTR [ecx+eax*4] ret 0 mov eax, DWORD PTR _n$[esp-4] mov ecx, DWORD PTR _A$[esp-4] mov eax, DWORD PTR [ecx+eax*4] ret 0
#include <iostream>#include <string>using namespace std;void alter(string names[]){ names[0] = "zombie"; names[1] = "vampire"; names[2] = "mummy"; names[3] = "werewolf"; names[4] = "grim reaper";}void main(void){ string names[] = {"imp", "critter", "slime", "goblin", "troll"}; int i=0; for(i=0;i<5;i++) { cout << names << endl; } cout << endl; alter(names); for(i=0;i<5;i++) { cout << names << endl; }}#include <iostream>#include <vector>#include <string>using namespace std;// turns out I find a use for this example below =)template< class T, int N >int array_size( T (&arg)[N] ) { return N; }// because we didnt pass the vector by reference, the// vector in main will not be affectedvoid alter(std::vector<std::string> names){ names[0] = "zombie"; names[1] = "vampire"; names[2] = "mummy"; names[3] = "werewolf"; names[4] = "grim reaper";}int main(){ const char *names_data[] = {"imp", "critter", "slime", "goblin", "troll"}; // there is no easy way to initialise a std::vector directly // currently this is about as easy as it gets // in practise though such data will be loaded from // an external source anyway std::vector<std::string> names(data, data + array_size(data); // declare loop variables inside the for loop // its much clearer //int i=0; for( int i = 0 ; i < names.size() ; i++ ) { cout << names << endl; } cout << endl; alter(names); for( int i = 0 ; i < names.size() ; i++ ) { cout << names << endl; }}Quote:Original post by Alpha Brain
but without the vector, it isn't possible? (just want to understand the basic way first)
