This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Sizeof an array in the C programming language?
#include "stdafx.h"
#include <string>
#include <iostream>
using namespace std;
string a[] = {"some", "text"};
void test(string a[])
{
int size_of_a = sizeof(a) /sizeof(a[0]);
cout << size_of_a;
}
int _tmain(int argc, _TCHAR* argv[])
{
test(a); //gives 0
int size_of_a = sizeof(a) /sizeof(a[0]);
cout << size_of_a; //gives 2
return 0;
}
as u can see in the comment test(a) gives 0 instead of 2 as i would expect. Could someone explain why and how could i correct it? thanks
When you pass an array to a function, it decays to a pointer to the first element of the array and so within your test(string a[]) function
sizeof(a);
actually returns the size of a pointer and not the size of your array.
To prevent array decaing to pointer, you can pass reference to array to the function. But it causes types of array of function formal argument and actual argument must coincide (including their sizes). So you need to use template to make your function work with an array of any size:
template <int N>
void foo(const string (&a)[N])
{
int size_of_a = sizeof(a) /sizeof(a[0]);
cout << size_of_a;
}
Related
This question already has answers here:
getting size of array from pointer c++
(6 answers)
Closed 1 year ago.
So I'm trying to make a function that gets the length of an array by returning the sizeof the array and sizeof the integer type...
code:
#include <cstdio>
#include <iostream>
using namespace std;
int len(int *thing) {
return sizeof(thing) / sizeof(int);
}
int main() {
int fard[] = {100, 45, 1, 723, 500};
cout << len(fard);
}
... and when I run the code it just returns with 1
how do I fix this / what did I do wrong.
The function parameter has a pointer type
int len(int *thing) {
return sizeof(thing) / sizeof(int);
}
Even if you will rewrite the function like
int len(int thing[]) {
return sizeof(thing) / sizeof(int);
}
nevertheless the compiler will adjust the array type of the function parameter to the type pointer to the array element type as written in the first function declaration.
So within the function the expression sizeof(thing) yields the size of a pointer. If the size of a pointer is equal to the value of sizeof( int ) then the function returns 1.
You could write instead
template <size_t N>
size_t len(const int ( &thing )[N] ) {
return N;
}
and then
cout << len(fard);
to get the number of elements in the array fard.
Pay attention to that there is already standard C++ function std::size declared in the header <iterator> in C++ 17.
So you could write
#include <iterator>
//...
std::cout << std::size(fard);
Because sizeof(thing) / sizeof(int) is probably equivalent to 4 / 4 == 1.
Because thing is a pointer and pointers have a size of 4 bytes on 32-bit compilers.
Here:
len(fard)
when you pass fard to the function, it decays to a pointer. This means that it loses its size information. So inside the function len you are dividing a single pointer by sizeof(int).
Try this instead
#include <cstdio>
#include <iostream>
using namespace std;
template< std::size_t N >
int len(int (&thing)[N] ) {
return sizeof(thing) / sizeof(int);
}
int main() {
int fard[] = {100, 45, 1, 723, 500};
cout << len(fard);
}
Result:
Program stdout
5
Code: https://godbolt.org/z/qzdbqEYx1
By the way, on x86_64 your original code returns 2 = sizeof(64 bit pointer)/sizeof(32 bit int)
This question already has answers here:
determine size of array if passed to function
(10 answers)
size of array passed to C++ function? [duplicate]
(7 answers)
Closed 3 years ago.
How do I get the size of bits of an Array from a function
int NumberOfElements(int Array[]);
int main()
{
int Array[] = { 5,5,6,5,5 };
std::cout << NumberOfElements(Array);
}
int NumberOfElements(int Array[]) {
return sizeof(Array);
}
It's returning 4.
Result should be 20.
Arrays decay into pointers when passed as arguments to functions etc.
The size 4 means that the pointer has that size. It does not tell you anything about the number of elements in the actual array.
You may want to use a std::vector<int> instead where the size is part of its interface:
#include <vector>
int main()
{
std::vector<int> Array{ 5,5,6,5,5 };
std::cout << NumberOfElements(Array);
}
int NumberOfElements(const std::vector<int>& Array) {
return Array.size();
}
This question already has answers here:
How to find the size of an array (from a pointer pointing to the first element array)?
(17 answers)
Passing Arrays to Function in C++
(5 answers)
What is array to pointer decay?
(11 answers)
Closed 9 years ago.
I made a function in C++ to find the length of an array. I find the sizeof the array passed in the argument and divide it by the sizeof the variable type. This should work but it always returns 1! Am I missing something obvious? Or does this have to do with pointers and memory? This is my code:
#include <iostream>
using namespace std;
int lengthOf(int arr[]);
int main() {
int x[] = {1,2,3,0,9,8};
int lenX = lengthOf(x);
cout << lenX;
return 0;
}
int lengthOf(int arr[]) {
int totalSize = sizeof arr;
cout << totalSize << endl;
int elementSize = sizeof(int);
return totalSize/elementSize;
}
Output (should be 6 instead of 1):
4
1
I am fairly new so excuse me if this is a bad question.
When passing an array as parameter, it always decays into a pointer. If you want to see the size of the array, you need to pass the array by reference, e.g.:
template <int Size>
int lengthOf(int (&array)[Size]) {
return Size;
}
You should use the pointer.
(sizeof(arr)/sizeof(*arr))
Even though int arr[] looks like you are passing an array, you are actually passing a pointer. int arr[] is equivalent to int* arr when used as a function parameter, this comes from C.
In C++, if you want to pass an array, the proper way is to do it by reference:
template <int N>
int lengthOf(int (&arr)[N]) {
int totalSize = sizeof arr;
cout << totalSize << endl;
int elementSize = sizeof(int);
return totalSize/elementSize;
}
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Passing an array as an argument in C++
Sizeof an array in the C programming language?
Can you please explain the output of the following code:
#include<iostream>
using namespace std;
void foo(int array[])
{
int size = sizeof(array) / sizeof(array[0]);
cout<<size<<endl;
}
int main()
{
int array[] = {1,2,3};
int size = sizeof(array) / sizeof(array[0]);
cout<<size<<endl;
foo(array);
return 0;
}
The corresponding output is:
3
2
Both the code inside foo() and inside main() looks similar to me so as to produce the same output, but it does not, can you please explain why?
void foo(int array[])
in C or C++ you cannot pass arrays by value, so the above declaration is interpretted as:
void foo(int * array)
Consider passing the array by reference:
template <size_t N>
void foo( int(&array)[N] )
{
cout << N << endl;
}
You are not passing the array into the function, you are passing the pointer to the array.
That means, in the function foo(), the result of sizeof(array) is the size of a pointer to an array of char, not the size of the array itself, so the function is effectively doing this:
cout << sizeof(int *) / sizeof(array[0]);
In this case size of int * is 8, size of int is 4, so you are getting an output of 2 from the function.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
2D arrays with C++
Hi, I'm trying to copy a pointer to a matrix that i'm passing in to a function in C++. here's what my code is trying to express
#include <iostream>
using namespace std;
void func( char** p )
{
char** copy = p;
cout << **copy;
}
int main()
{
char x[5][5];
x[0][0] = 'H';
func( (char**) &x);
return 0;
}
However, this gives me a Seg Fault. Would someone please explain (preferrably in some detail) what underlying mechanism i'm missing out on? (and the fix for it)
Thanks much in advance :)
A pointer to an array of 5 arrays of 5 char (char x[5][5]) has the type "pointer to array of 5 arrays of 5 chars", that is char(*p)[5][5]. The type char** has nothing to do with this.
#include <iostream>
using namespace std;
void func( char (*p)[5][5] )
{
char (*copy)[5][5] = p;
cout << (*copy)[0][0];
}
int main()
{
char x[5][5];
x[0][0] = 'H';
func(&x);
return 0;
}
Of course there are many other ways to pass a 2D array by reference or pointer, as already mentioned in comments. The most in-detail reference is probably StackOverflow's own C++ FAQ, How do I use arrays in C++?
char** is a pointer to a pointer (or an array of pointers). &x is not one of those - it's a pointer to a two-dimensional array of chars, which can be implicitly converted to a pointer to a single char (char *). The compiler probably gave you an error, at which point you put in the cast, but the compiler was trying to tell you something important.
Try this instead of using a char**:
#include <iostream>
using namespace std;
void func( char* &p )
{
char* copy = p;
cout << copy[0];
}
int main()
{
char x[5][5];
x[0][0] = 'H';
func(&x[0]);
return 0;
}