I'm teaching myself C++ and had some questions about arrays and pointers. My understanding is that arrays are really just pointers, however, arrays are address constants which cannot be changed.
If this is the case, I was wondering why in my function show2() I was able to change the address of the pointer list. Unlike variables, I thought arrays are passed by reference so I was expecting a compiler error when calling function show2() since I incremented the address of list. But the code works just fine. Can someone please explain?
Thank you!
#include<iostream>
#include<iomanip>
using namespace std;
void show1(double *list, int SIZE)
{
for(int i=0; i < SIZE; i++)
{
cout << setw(5) << *(list+i);
}
cout << endl;
return;
}
void show2(double *list, int SIZE)
{
double *ptr = list;
for(int i=0; i < SIZE; i++)
cout << setw(5) << *list++;
cout << endl;
return;
}
int main()
{
double rates[] = {6.5, 7.2, 7.5, 8.3, 8.6,
9.4, 9.6, 9.8, 10.0};
const int SIZE = sizeof(rates) / sizeof(double);
show1(rates, SIZE);
show2(rates, SIZE);
return 0;
}
My understanding is that arrays are really just pointers
Let's get that out of the way. No, arrays are not pointers. Arrays are a series of objects, all of the same type, contiguous in memory.
Arrays can be passed by reference, but that is not what is usually done. What is usually done, which is what you are doing, is passing a pointer to the first element of the array. Arrays can and will "decay" to a pointer to their first element upon demand. And that's what is happening when you pass rates to show1 and show2.
Inside show1 and show2, list starts out as a pointer to rates[0]. You're free to modify this pointer to point at any other double.
If you wanted to pass an array by reference, it would look like this:
void show3(double (&list)[9]) { ... }
Or the more versatile:
template<size_t SIZE>
void show3(double (&list)[SIZE]) { ... }
Note that what you can't do is pass an array by value (unless it is contained within another object). If you write a function which looks like it is taking an array by value, e.g.
void show4(double list[9]) { ... }
It is actually a pointer, and that number 9 is meaningless. Native arrays suck.
First, arrays are converted to a pointer to the first element when passed as the function argument. BTW, arrays are not pointers, as one example, sizeof(rates) in your code isn't the size of a pointer.
Second, arrays are passed by value since you are not using references.
So in the function show2, you are modifying a pointer, which is fine.
Arrays are not pointers. C++ has inherited "Array-Pointer Equivalence" from C which means that a well-known array variable can decay to a pointer, primarily for the purpose of offset math and for avoiding passing arrays by value:
int array[64];
int* a = array; // equivalent to a = &array[0];
Array's aren't pointers. If you use an array variable name in a pointer context, it will "decay" to a pointer - that is, lose the extended attributes available from an array object.
int array[64];
int* a = array;
std::cout << "array size = " << sizeof(array) << "\n";
std::cout << "a size = " << sizeof(a) << "\n";
std::cout << "(int*)(array) size = " << sizeof((int*)array)) << "\n";
"Array size" will be 256 (int is 4 bytes, 64 of them = 256 bytes), "a size" will be 4 or 8 bytes depending on 32/64 bits, and "(int*)(array)" size will be the same size as the pointer.
People often think that arrays are passed by value. This is not true: http://ideone.com/hAeH18
#include <iostream>
void bump(int arr[3]) {
for (size_t i = 0; i < 3; ++i)
arr[i]++;
}
int main() {
int array[] = { 1, 2, 3 };
bump(array);
for (size_t i = 0; i < 3; ++i)
std::cout << array[i] << "\n";
return 0;
}
This outputs "2, 3, 4" not "1, 2, 3".
This occurs because arrays decay to pointers when passed as function arguments. But to support the syntax for receiving arrays as arrays, C has to be able to treat pointers like arrays in some contexts:
void f1(int* a) { a[0]++; }
void f2(int* a) { (*a)++; }
void f3(int a[]) { a[0]++; }
void f4(int a[]) { (*a)++; }
void f5(int a[1]) { a[0]++; }
void f6(int a[1]) { (*a)++; }
All of these functions produce the same code.
In C, this originated from the fact that array information is lost at compile time. So this function:
void f(int array[])
has no way to tell how large the array it is receiving is. They wanted programmers to be conscious of this and be careful about how/if they passed size information - e.g. in the case of char arrays, instead of size, we have the nul terminator byte.
Unfortunately they didn't choose to make it obvious by diasllowing the representation that makes it look like you are receiving an array with size information intact :(
Related
I am writing a simple function that returns the largest integer in an array. The problem I am having is finding the number of elements in the array.
Here is the function header:
int largest(int *list, int highest_index)
How can I get the number of integers in the array 'list'.
I have tried the following methods:
int i = sizeof list/sizeof(int); //returns incorrect value
int i = list.size(); // does not compile
Any help will be greatly appreciated!
C++ is based on C and inherits many features from it. In relation to this question, it inherits something called "array/pointer equivalence" which is a rule that allows an array to decay to a pointer, especially when being passed as a function argument. It doesn't mean that an array is a pointer, it just means that it can decay to one.
void func(int* ptr);
int array[5];
int* ptr = array; // valid, equivalent to 'ptr = &array[0]'
func(array); // equivalent to func(&array[0]);
This last part is the most relevant to your question. You are not passing the array, you are passing the address of the 0th element.
In order for your function to know how big the incoming array is, you will need to send that information as an argument.
static const size_t ArraySize = 5;
int array[ArraySize];
func(array, ArraySize);
Because the pointer contains no size information, you can't use sizeof.
void func(int* array) {
std::cout << sizeof(array) << "\n";
}
This will output the size of "int*" - which is 4 or 8 bytes depending on 32 vs 64 bits.
Instead you need to accept the size parameters
void func(int* array, size_t arraySize);
static const size_t ArraySize = 5;
int array[ArraySize];
func(array, ArraySize);
Even if you try to pass a fixed-sized array, it turns out this is syntactic sugar:
void func(int array[5]);
http://ideone.com/gaSl6J
Remember how I said that an array is NOT a pointer, just equivalent?
int array[5];
int* ptr = array;
std::cout << "array size " << sizeof(array) << std::endl;
std::cout << "ptr size " << sizeof(ptr) << str::endl;
array size will be 5 * sizeof(int) = 20
ptr size will be sizeof(int *) which will be either 4 or 8 bytes.
sizeof returns the size of the type being supplied, if you supply an object then it deduces the type and returns the size of that
If you want to know how many elements of an array are in the array, when you have the array and not a pointer, you can write
sizeof(array) / sizeof(array[0])
or
sizeof(array) / sizeof(*array)
There's no way to do that. This is one good reason (among many) to use vectors instead of arrays. But if you must use an array then you must pass the size of the array as a parameter to your function
int largest(int *list, int list_size, int highest_index)
Arrays in C++ are quite poor, the sooner you learn to use vectors the easier you will find things.
Pointers do not have information about the number of elements they refer to. If you are speaking about the first argument of the function call then if list is an array you can indeed use the syntax
sizeof( list ) / sizeof( int )
I would like to append that there are three approaches. The first one is to use arrays passed by reference. The second one is to use pointer to the first element and the number of elements. And the third one is to use two pointers - the start pointer and the last pointer as standard algorithms usually are defined. Character arrays have an additional possibility to process them.
The simple answer is you cannot. You need to store it in a variable. The great advantage with C++ is it has STL and you can use vector. size() method gives the size of the vector at that instant.
#include<iostream>
#include<vector>
using namespace std;
int main () {
vector<int> v;
for(int i = 0; i < 10; i++) {
v.push_back(i);
}
cout << v.size() << endl;
for(int i = 0; i < 10; i++) {
v.push_back(i);
}
cout << v.size() << endl;
return 0;
}
output:
10
20
Not tested. But, should work. ;)
You need to remember in a variable array size, there is no possibility to retrieve array size from pointer.
const int SIZE = 10;
int list[SIZE];
// or
int* list = new int[SIZE]; // do not forget to delete[]
My answer uses a char array instead of an integer array but I do hope it helps.
You can use a counter until you reach the end of the array. Char arrays always end with a '\0' and you can use that to check the end of the array is reached.
char array[] = "hello world";
char *arrPtr = array;
char endOfString = '\0';
int stringLength = 0;
while (arrPtr[stringLength] != endOfString) {
stringLength++;
}
stringLength++;
cout << stringLength << endl;
Now you have the length of the char array.
I tried the counting the number of integers in array using this method. Apparently, '\0' doesn't apply here obviously but the -1 index of the array is 0. So assuming that there is no 0, in the array that you are using. You can replace the '\0' with 0 in the code and change the code to use int pointers and arrays.
I am writing a simple function that returns the largest integer in an array. The problem I am having is finding the number of elements in the array.
Here is the function header:
int largest(int *list, int highest_index)
How can I get the number of integers in the array 'list'.
I have tried the following methods:
int i = sizeof list/sizeof(int); //returns incorrect value
int i = list.size(); // does not compile
Any help will be greatly appreciated!
C++ is based on C and inherits many features from it. In relation to this question, it inherits something called "array/pointer equivalence" which is a rule that allows an array to decay to a pointer, especially when being passed as a function argument. It doesn't mean that an array is a pointer, it just means that it can decay to one.
void func(int* ptr);
int array[5];
int* ptr = array; // valid, equivalent to 'ptr = &array[0]'
func(array); // equivalent to func(&array[0]);
This last part is the most relevant to your question. You are not passing the array, you are passing the address of the 0th element.
In order for your function to know how big the incoming array is, you will need to send that information as an argument.
static const size_t ArraySize = 5;
int array[ArraySize];
func(array, ArraySize);
Because the pointer contains no size information, you can't use sizeof.
void func(int* array) {
std::cout << sizeof(array) << "\n";
}
This will output the size of "int*" - which is 4 or 8 bytes depending on 32 vs 64 bits.
Instead you need to accept the size parameters
void func(int* array, size_t arraySize);
static const size_t ArraySize = 5;
int array[ArraySize];
func(array, ArraySize);
Even if you try to pass a fixed-sized array, it turns out this is syntactic sugar:
void func(int array[5]);
http://ideone.com/gaSl6J
Remember how I said that an array is NOT a pointer, just equivalent?
int array[5];
int* ptr = array;
std::cout << "array size " << sizeof(array) << std::endl;
std::cout << "ptr size " << sizeof(ptr) << str::endl;
array size will be 5 * sizeof(int) = 20
ptr size will be sizeof(int *) which will be either 4 or 8 bytes.
sizeof returns the size of the type being supplied, if you supply an object then it deduces the type and returns the size of that
If you want to know how many elements of an array are in the array, when you have the array and not a pointer, you can write
sizeof(array) / sizeof(array[0])
or
sizeof(array) / sizeof(*array)
There's no way to do that. This is one good reason (among many) to use vectors instead of arrays. But if you must use an array then you must pass the size of the array as a parameter to your function
int largest(int *list, int list_size, int highest_index)
Arrays in C++ are quite poor, the sooner you learn to use vectors the easier you will find things.
Pointers do not have information about the number of elements they refer to. If you are speaking about the first argument of the function call then if list is an array you can indeed use the syntax
sizeof( list ) / sizeof( int )
I would like to append that there are three approaches. The first one is to use arrays passed by reference. The second one is to use pointer to the first element and the number of elements. And the third one is to use two pointers - the start pointer and the last pointer as standard algorithms usually are defined. Character arrays have an additional possibility to process them.
The simple answer is you cannot. You need to store it in a variable. The great advantage with C++ is it has STL and you can use vector. size() method gives the size of the vector at that instant.
#include<iostream>
#include<vector>
using namespace std;
int main () {
vector<int> v;
for(int i = 0; i < 10; i++) {
v.push_back(i);
}
cout << v.size() << endl;
for(int i = 0; i < 10; i++) {
v.push_back(i);
}
cout << v.size() << endl;
return 0;
}
output:
10
20
Not tested. But, should work. ;)
You need to remember in a variable array size, there is no possibility to retrieve array size from pointer.
const int SIZE = 10;
int list[SIZE];
// or
int* list = new int[SIZE]; // do not forget to delete[]
My answer uses a char array instead of an integer array but I do hope it helps.
You can use a counter until you reach the end of the array. Char arrays always end with a '\0' and you can use that to check the end of the array is reached.
char array[] = "hello world";
char *arrPtr = array;
char endOfString = '\0';
int stringLength = 0;
while (arrPtr[stringLength] != endOfString) {
stringLength++;
}
stringLength++;
cout << stringLength << endl;
Now you have the length of the char array.
I tried the counting the number of integers in array using this method. Apparently, '\0' doesn't apply here obviously but the -1 index of the array is 0. So assuming that there is no 0, in the array that you are using. You can replace the '\0' with 0 in the code and change the code to use int pointers and arrays.
I am writing a simple function that returns the largest integer in an array. The problem I am having is finding the number of elements in the array.
Here is the function header:
int largest(int *list, int highest_index)
How can I get the number of integers in the array 'list'.
I have tried the following methods:
int i = sizeof list/sizeof(int); //returns incorrect value
int i = list.size(); // does not compile
Any help will be greatly appreciated!
C++ is based on C and inherits many features from it. In relation to this question, it inherits something called "array/pointer equivalence" which is a rule that allows an array to decay to a pointer, especially when being passed as a function argument. It doesn't mean that an array is a pointer, it just means that it can decay to one.
void func(int* ptr);
int array[5];
int* ptr = array; // valid, equivalent to 'ptr = &array[0]'
func(array); // equivalent to func(&array[0]);
This last part is the most relevant to your question. You are not passing the array, you are passing the address of the 0th element.
In order for your function to know how big the incoming array is, you will need to send that information as an argument.
static const size_t ArraySize = 5;
int array[ArraySize];
func(array, ArraySize);
Because the pointer contains no size information, you can't use sizeof.
void func(int* array) {
std::cout << sizeof(array) << "\n";
}
This will output the size of "int*" - which is 4 or 8 bytes depending on 32 vs 64 bits.
Instead you need to accept the size parameters
void func(int* array, size_t arraySize);
static const size_t ArraySize = 5;
int array[ArraySize];
func(array, ArraySize);
Even if you try to pass a fixed-sized array, it turns out this is syntactic sugar:
void func(int array[5]);
http://ideone.com/gaSl6J
Remember how I said that an array is NOT a pointer, just equivalent?
int array[5];
int* ptr = array;
std::cout << "array size " << sizeof(array) << std::endl;
std::cout << "ptr size " << sizeof(ptr) << str::endl;
array size will be 5 * sizeof(int) = 20
ptr size will be sizeof(int *) which will be either 4 or 8 bytes.
sizeof returns the size of the type being supplied, if you supply an object then it deduces the type and returns the size of that
If you want to know how many elements of an array are in the array, when you have the array and not a pointer, you can write
sizeof(array) / sizeof(array[0])
or
sizeof(array) / sizeof(*array)
There's no way to do that. This is one good reason (among many) to use vectors instead of arrays. But if you must use an array then you must pass the size of the array as a parameter to your function
int largest(int *list, int list_size, int highest_index)
Arrays in C++ are quite poor, the sooner you learn to use vectors the easier you will find things.
Pointers do not have information about the number of elements they refer to. If you are speaking about the first argument of the function call then if list is an array you can indeed use the syntax
sizeof( list ) / sizeof( int )
I would like to append that there are three approaches. The first one is to use arrays passed by reference. The second one is to use pointer to the first element and the number of elements. And the third one is to use two pointers - the start pointer and the last pointer as standard algorithms usually are defined. Character arrays have an additional possibility to process them.
The simple answer is you cannot. You need to store it in a variable. The great advantage with C++ is it has STL and you can use vector. size() method gives the size of the vector at that instant.
#include<iostream>
#include<vector>
using namespace std;
int main () {
vector<int> v;
for(int i = 0; i < 10; i++) {
v.push_back(i);
}
cout << v.size() << endl;
for(int i = 0; i < 10; i++) {
v.push_back(i);
}
cout << v.size() << endl;
return 0;
}
output:
10
20
Not tested. But, should work. ;)
You need to remember in a variable array size, there is no possibility to retrieve array size from pointer.
const int SIZE = 10;
int list[SIZE];
// or
int* list = new int[SIZE]; // do not forget to delete[]
My answer uses a char array instead of an integer array but I do hope it helps.
You can use a counter until you reach the end of the array. Char arrays always end with a '\0' and you can use that to check the end of the array is reached.
char array[] = "hello world";
char *arrPtr = array;
char endOfString = '\0';
int stringLength = 0;
while (arrPtr[stringLength] != endOfString) {
stringLength++;
}
stringLength++;
cout << stringLength << endl;
Now you have the length of the char array.
I tried the counting the number of integers in array using this method. Apparently, '\0' doesn't apply here obviously but the -1 index of the array is 0. So assuming that there is no 0, in the array that you are using. You can replace the '\0' with 0 in the code and change the code to use int pointers and arrays.
To fully understand how pointers, values, and references work, I am making a basic C++ program that attempts to tamper with some static and dynamic arrays and understand exactly how they should be passed in.
First I generate a static array of 3 elements. I then pass it into a function that modifies all elements. I then pass it into another function with a slightly different signature, but can also alter the array's values.
Next I generate a dynamically sized array, pass it into a function by reference so that all of the values in this dynamically sized array can be altered.
The code is as follows:
#include "stdafx.h"
#include <iostream>
#include <string>
using namespace std;
void changeIndexStaticArrayMethod1(int* stat);
void changeIndexStaticArrayMethod2(int (&stat)[3]);
void changeIndexDynamicArrayMethod1(int* dyn, int size);
int main() {
const int MAX = 3;
int arr[MAX] = { 1,2,3 };
changeIndexStaticArrayMethod1(arr);
cout << arr[0] << endl;
cout << arr[1] << endl;
cout << arr[2] << endl;
cout << endl;
changeIndexStaticArrayMethod2(arr);
cout << arr[0] << endl;
cout << arr[1] << endl;
cout << arr[2] << endl;
int SIZE;
cout << "Please choose a size for the array" << endl;
cin >> SIZE;
int *ne = new int[SIZE];
//Build array
for (int i = 0; i < SIZE; i++) {
ne[i] = i;
}
changeIndexDynamicArrayMethod1(ne, SIZE);
for (int i = 0; i < SIZE; i++) {
cout << "ne[" << i << "] = " << ne[i] << endl;
}
//To hang program
cin >> SIZE;
delete[] arr;
delete[] ne;
return 0;
}
void changeIndexStaticArrayMethod1(int* stat) {
stat[0] = 10;
stat[1] = 20;
stat[2] = 30;
}
void changeIndexStaticArrayMethod2(int (&stat)[3]) {
stat[0] = 40;
stat[1] = 50;
stat[2] = 60;
}
void changeIndexDynamicArrayMethod1(int* dyn, int size) {
for (int i = 0; i < size; i++) {
dyn[i] = i * 10;
}
}
All of the above code works how I want it to, I just have a few questions as to why (some of the methods of passing arrays by reference I have found on other SO questions).
In the changeIndexStaticArrayMethod1() and changeIndexDynamicArrayMethod1() functions, why are we able to use the dereference * operator for our array as reference? My knee jerk reaction is seeing that as practically passing the array in by values since it is the dereference operator. I know that with arrays, it is much different than using variables, but also, why will the following not work for single int variables:
void changeStaticNumber(int* num){
num = 100;
}
Obviously the above will work if we use &num and not int* num, and obviously I don't fully understand the relationship between pointers and arrays, but I cannot figure out why when we pass an array by reference, int* staticArray is ok.
Any explanation for these problems I am having would be much appreciated. Thanks.
why are we able to use the dereference * operator for our array as reference?
The * in C means many things. It can mean the unary indirection ("contents of") operator, it can mean the binary multiplication operator and it can mean a pointer declaration. The int* stat is a pointer declaration.
Since you aren't using the * to dereference the contents of the pointer inside that function, I'm not quite sure what you are asking.
When you take the array name of your array in main(), it "decays" into a pointer to the first element. So what those function do, is to take a pointer by value. If you dereference the pointer by typing *stat = something; you access the actual array in main.
Should you do something weird like changing the pointer itself, for example stat++;, then it will not affect the address used in main. You passed the pointer itself by value, so the pointer is a local copy.
My knee jerk reaction is seeing that as practically passing the array in by values since it is the dereference operator.
You can't really pass arrays by value in C or C++, without resorting to dirty tricks (storing them inside structs or classes). For example, had your function been written as void changeIndexStaticArrayMethod1(int stat[3]) it would still give you a pointer to the first element. It will not pass an array by value, as the syntax might trick you into believing.
why will the following not work for single int variables:
void changeStaticNumber(int* num){ num = 100; }
Because num is the pointer itself, not its contents. In order to write code like that, you could pass the variable by reference int& num. Behind the lines this is really the same thing as passing a pointer, just with simplified syntax.
To understand the relation between pointers and arrays better, start by reading this whole chapter: http://c-faq.com/aryptr/index.html (C and C++ are identical when it comes to pointers).
Let me see if I can take a stab at this.
Pointers are simply address holders. Once you do int * ptr = myarray; --- what you are in tern doing is storing the address of the pointer my array into ptr --- array names are actually pointers to the first memory location in the array. You can use pointer arithmetic to get at everything else for example myarray +1 will point you to the next location or myarray[1].
Passing by value is not very useful when you need to modify your array. Passing in by reference is essentially making a pointer to the array and passing that. Since arrays like vectors are contiguous blocks of memory you can index through them rather easily.
As far as your example goes void changeStaticNumber(int* num){ num = 100; } will not work because what you are attempting to do is store 100 into the pointer's address. If you deference num and make it void changeStaticNumber(int* num){ *num = 100; } it will work because you are actually going one step further and accessing the data that num is pointing to. When you use &num it is essentially the same thing - & just gives you the address of something.
For example if you want to point a pointer to an int what you would do is
int num = 5;
int *ptr = #
at this point in time ptr has the same address in num. To print out the data in num or that ptr is pointing to you need to dereference or go one step further as I like to tell myself and dereference to so cout << *ptr;
In both changeIndexStaticArrayMethod1 and changeIndexDynamicArrayMethod1 you are not passing an array there is no pass by reference (which only happens if the parameter type is a reference type -- i.e. with &). The parameter has type int * (pointer to int). You are passing a pointer to int by value. There is no "dereference operator" in either function.
ne is already an int *, so passing it is nothing special. arr is an int [3], an array, not a pointer. In C, when an array-of-T is used in a context that expects a pointer-to-T, it is implicitly converted (without you needing to do anything) to a pointer to its first element. So when you do, changeIndexStaticArrayMethod1(arr), the compiler gets a pointer to the first element of arr, and passes that to the function.
The [] operator works on pointers. a[i] is always guaranteed to be the same as *(a + i). Inside both the changeIndexStaticArrayMethod1 and changeIndexDynamicArrayMethod1 functions, [] is used to access subsequent elements using a pointer to the first element.
I am writing a simple function that returns the largest integer in an array. The problem I am having is finding the number of elements in the array.
Here is the function header:
int largest(int *list, int highest_index)
How can I get the number of integers in the array 'list'.
I have tried the following methods:
int i = sizeof list/sizeof(int); //returns incorrect value
int i = list.size(); // does not compile
Any help will be greatly appreciated!
C++ is based on C and inherits many features from it. In relation to this question, it inherits something called "array/pointer equivalence" which is a rule that allows an array to decay to a pointer, especially when being passed as a function argument. It doesn't mean that an array is a pointer, it just means that it can decay to one.
void func(int* ptr);
int array[5];
int* ptr = array; // valid, equivalent to 'ptr = &array[0]'
func(array); // equivalent to func(&array[0]);
This last part is the most relevant to your question. You are not passing the array, you are passing the address of the 0th element.
In order for your function to know how big the incoming array is, you will need to send that information as an argument.
static const size_t ArraySize = 5;
int array[ArraySize];
func(array, ArraySize);
Because the pointer contains no size information, you can't use sizeof.
void func(int* array) {
std::cout << sizeof(array) << "\n";
}
This will output the size of "int*" - which is 4 or 8 bytes depending on 32 vs 64 bits.
Instead you need to accept the size parameters
void func(int* array, size_t arraySize);
static const size_t ArraySize = 5;
int array[ArraySize];
func(array, ArraySize);
Even if you try to pass a fixed-sized array, it turns out this is syntactic sugar:
void func(int array[5]);
http://ideone.com/gaSl6J
Remember how I said that an array is NOT a pointer, just equivalent?
int array[5];
int* ptr = array;
std::cout << "array size " << sizeof(array) << std::endl;
std::cout << "ptr size " << sizeof(ptr) << str::endl;
array size will be 5 * sizeof(int) = 20
ptr size will be sizeof(int *) which will be either 4 or 8 bytes.
sizeof returns the size of the type being supplied, if you supply an object then it deduces the type and returns the size of that
If you want to know how many elements of an array are in the array, when you have the array and not a pointer, you can write
sizeof(array) / sizeof(array[0])
or
sizeof(array) / sizeof(*array)
There's no way to do that. This is one good reason (among many) to use vectors instead of arrays. But if you must use an array then you must pass the size of the array as a parameter to your function
int largest(int *list, int list_size, int highest_index)
Arrays in C++ are quite poor, the sooner you learn to use vectors the easier you will find things.
Pointers do not have information about the number of elements they refer to. If you are speaking about the first argument of the function call then if list is an array you can indeed use the syntax
sizeof( list ) / sizeof( int )
I would like to append that there are three approaches. The first one is to use arrays passed by reference. The second one is to use pointer to the first element and the number of elements. And the third one is to use two pointers - the start pointer and the last pointer as standard algorithms usually are defined. Character arrays have an additional possibility to process them.
The simple answer is you cannot. You need to store it in a variable. The great advantage with C++ is it has STL and you can use vector. size() method gives the size of the vector at that instant.
#include<iostream>
#include<vector>
using namespace std;
int main () {
vector<int> v;
for(int i = 0; i < 10; i++) {
v.push_back(i);
}
cout << v.size() << endl;
for(int i = 0; i < 10; i++) {
v.push_back(i);
}
cout << v.size() << endl;
return 0;
}
output:
10
20
Not tested. But, should work. ;)
You need to remember in a variable array size, there is no possibility to retrieve array size from pointer.
const int SIZE = 10;
int list[SIZE];
// or
int* list = new int[SIZE]; // do not forget to delete[]
My answer uses a char array instead of an integer array but I do hope it helps.
You can use a counter until you reach the end of the array. Char arrays always end with a '\0' and you can use that to check the end of the array is reached.
char array[] = "hello world";
char *arrPtr = array;
char endOfString = '\0';
int stringLength = 0;
while (arrPtr[stringLength] != endOfString) {
stringLength++;
}
stringLength++;
cout << stringLength << endl;
Now you have the length of the char array.
I tried the counting the number of integers in array using this method. Apparently, '\0' doesn't apply here obviously but the -1 index of the array is 0. So assuming that there is no 0, in the array that you are using. You can replace the '\0' with 0 in the code and change the code to use int pointers and arrays.