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How to print size of array parameter in a function in C++?
The size of a data type can be obtained using sizeof(). A program that demonstrates the printing of the array parameter in a function in C++ is given as follows.
Example
#include <iostream>
using namespace std;
int func(int a[]) {
cout << "Size: " << sizeof(a);
return 0;
}
int main() {
int array[5];
func(array);
cout << "\nSize: " << sizeof(array);
return 0;
}Output
The output of the above program is as follows.
Size: 8 Size: 20
Now let us understand the above program.
In the function func(), the size of a is displayed which is 8 because the array in main() is passed as a pointer and a points to the start of array. So, sizeof(a) displays the size of the pointer which is 8. The code snippet that shows this is as follows.
int func(int a[]) {
cout << "Size: " << sizeof(a);
return 0;
}In the function main(), the size of array is displayed which is 20. This is because the size of int is 4 and the array contains 5 int elements. The code snippet that shows this is as follows.
int main() {
int array[5];
func(array);
cout << "\nSize: " << sizeof(array);
return 0;
}
Updated on: 26-Jun-2020
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