If you have your array in scope you can use sizeof to determine its size in bytes and use the division to calculate the number of elements:
#define NUM_OF_ELEMS 10
int arr[NUM_OF_ELEMS];
size_t NumberOfElements = sizeof(arr)/sizeof(arr[0]);
If you receive an array as a function argument or allocate an array in heap you can not determine its size using the sizeof. You'll have to store/pass the size information somehow to be able to use it:
void DoSomethingWithArray(int* arr, int NumOfElems)
{
for(int i = 0; i < NumOfElems; ++i) {
arr[i] = /*...*/
}
}
If you have your array in scope you can use sizeof to determine its size in bytes and use the division to calculate the number of elements:
#define NUM_OF_ELEMS 10
int arr[NUM_OF_ELEMS];
size_t NumberOfElements = sizeof(arr)/sizeof(arr[0]);
If you receive an array as a function argument or allocate an array in heap you can not determine its size using the sizeof. You'll have to store/pass the size information somehow to be able to use it:
void DoSomethingWithArray(int* arr, int NumOfElems)
{
for(int i = 0; i < NumOfElems; ++i) {
arr[i] = /*...*/
}
}
int a[20];
int length;
length = sizeof(a) / sizeof(int);
and you can use another way to make your code not be hard-coded to int
Say if you have an array array
you just need to:
int len = sizeof(array) / sizeof(array[0]);
hi, I'm new in C. I'm having a hard time finding the number of elements in an int array.
Here's what I'm trying to achieve.
I've defined a macro#define MAXSZ 30
Then I've declared an arrayint arr[MAXSZ];
What I wanted to do, is, taking user input to initialize the array up until the user inputs -1. Then get the number of elements.int item;for(int i = 0; i < MAXSZ; i++) {scanf("%d", &item);if (item == -1) break;else arr[i] = item;}
Now, when I'm trying to get the number of elements.
For input12356-1
It shows 30 elements (as it's supposed to).
Output300x7ffe9be05320 -> 10x7ffe9be05324 -> 20x7ffe9be05328 -> 30x7ffe9be0532c -> 50x7ffe9be05330 -> 60x7ffe9be05334 -> 325720x7ffe9be05338 -> -12939427840x7ffe9be0533c -> -1823360x7ffe9be05340 -> -13812227900x7ffe9be05344 -> 325720x7ffe9be05348 -> 0
...
My question is, how do I get rid of the garbage values & only get the number of items that has been given as input?
Count the number of elements in an array in C - Stack Overflow
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C: finding the number of elements in an array[] - Stack Overflow
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In C, you can only get the size of statically allocated arrays, i.e.
int array[10];
size = sizeof(array) / sizeof(int);
would give 10.
If your array is declared or passed as int* array, there is no way to determine its size, given this pointer only.
You are most likely doing this inside the function to which you pass the array.
The array decays as pointer to first element So You cannot do so inside the called function.
Perform this calculation before calling the function and pass the size as an function argument.
Executive summary:
int a[17];
size_t n = sizeof(a)/sizeof(a[0]);
Full answer:
To determine the size of your array in bytes, you can use the sizeof
operator:
int a[17];
size_t n = sizeof(a);
On my computer, ints are 4 bytes long, so n is 68.
To determine the number of elements in the array, we can divide the total size of the array by the size of the array element. You could do this with the type, like this:
int a[17];
size_t n = sizeof(a) / sizeof(int);
and get the proper answer (68 / 4 = 17), but if the type of
a changed you would have a nasty bug if you forgot to change
the sizeof(int) as well.
So the preferred divisor is sizeof(a[0]) or the equivalent sizeof(*a), the size of the first element of the array.
int a[17];
size_t n = sizeof(a) / sizeof(a[0]);
Another advantage is that you can now easily parameterize the array name in a macro and get:
#define NELEMS(x) (sizeof(x) / sizeof((x)[0]))
int a[17];
size_t n = NELEMS(a);
The sizeof way is the right way iff you are dealing with arrays not received as parameters. An array sent as a parameter to a function is treated as a pointer, so sizeof will return the pointer's size, instead of the array's.
Thus, inside functions this method does not work. Instead, always pass an additional parameter size_t size indicating the number of elements in the array.
Test:
#include <stdio.h>
#include <stdlib.h>
void printSizeOf(int intArray[]);
void printLength(int intArray[]);
int main(int argc, char* argv[])
{
int array[] = { 0, 1, 2, 3, 4, 5, 6 };
printf("sizeof of array: %d\n", (int) sizeof(array));
printSizeOf(array);
printf("Length of array: %d\n", (int)( sizeof(array) / sizeof(array[0]) ));
printLength(array);
}
void printSizeOf(int intArray[])
{
printf("sizeof of parameter: %d\n", (int) sizeof(intArray));
}
void printLength(int intArray[])
{
printf("Length of parameter: %d\n", (int)( sizeof(intArray) / sizeof(intArray[0]) ));
}
Output (in a 64-bit Linux OS):
sizeof of array: 28
sizeof of parameter: 8
Length of array: 7
Length of parameter: 2
Output (in a 32-bit windows OS):
sizeof of array: 28
sizeof of parameter: 4
Length of array: 7
Length of parameter: 1