CODEWITHSUNDEEP
carrays
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Debugger

Reputation: 31

What is the result of the code?

please interpret for me about the result that the following code run.

#include <stdio.h>

int get_array_size(int *arr){
    return sizeof(arr)/sizeof(int);
}

int main(int argc, char const *argv[]){
    int arr[10];
    printf("%d\n", sizeof(arr)/sizeof(int));
    printf("%d\n",get_array_size(arr));
    printf("%d\n",arr[10]);
    printf("%d\n",get_array_size(arr));
    return 0;
}

which is:

10                                                                                                                                                                                                   
2                                                                                                                                                                                                    
0                                                                                                                                                                                                    
2

Upvotes: 0

Views: 62

Answers (3)

MayurK
MayurK

Reputation: 1957

printf("%d\n", sizeof(arr)/sizeof(int));

sizeof(arr) returns size of array in bytes. Since it is an int array of length 10, "sizeof(arr)" will be "10*sizeof(int)". You are deviding it by sizeof(int). So output will be 10.

printf("%d\n",get_array_size(arr));

This is interesting! You are passing address of "arr" to get_array_size(). In that function first you are getting size of address. The address size depends on the machine you are running, i.e. In 32-bit machine address size will be 4bytes and in 64-bit machine it will be 8bytes. I think you are using 64-bit machine so sizeof(arr) is returning 8. But sizeof(int) is 4bytes. So effectively it is 8/4 = 2.

printf("%d\n",arr[10]);

Your array size is 10, so you can access from arr[0] to arr[9]. arr[10] is not part of your array. That address might be used for some other variable. You are just writing whatever present in that address. The output is completely random. I think that address had 0 when you executed the program, so you got the output as 0.

printf("%d\n",get_array_size(arr));

This is repeated print (Output is 2).

Upvotes: 1

Anjaneyulu
Anjaneyulu

Reputation: 434

#include <stdio.h>

int get_array_size(int *arr){
    return sizeof(arr)/sizeof(int);
}

int main(int argc, char const *argv[]){
    int arr[10];
    printf("%d\n", sizeof(arr)/sizeof(int));//10
    printf("%d\n",get_array_size(arr));//1
    printf("%d\n",arr[10]);//if a[9] adress=x a[10] prints undefined value from x+4 adress
    printf("%d\n",get_array_size(arr));1
    return 0;
}

When you call the function local variable arr is given priority so 4/4 always gives 1 when called

Upvotes: 1

v78
v78

Reputation: 2933

All sizes are in bytes.

sizeof(arr)/sizeof(int) = size of array / size of int = 40/4 = 10 

get_array_size(arr) = size of array pointer(which is 4) / size of int = 4/4 = 1. 

arr[10] = the eleventh entry of array(array size=10). (undefined number).

Upvotes: 1

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