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Even though "a" is declared as an array of 10 bytes, "a" is actually a pointer. The function is executed on a 16 bit machine.

a is pointer Which occupy 2 byte so the out put is 2

array name is pointer . and every pointer's size is integer

i dont think so when you run this code char a[10]; printf("%d",sizeof(a)); this will give you 40 that is actual size of array. one explanation could be . function does not accept array as a whole its just accepts pointers or any other contents as a parameter and in this case a is a pointer not array hence 2

when we pass any array in c it actually decays into the pointer to the first member of array in this case that is integer and u are running it in 16 bit machine that is why the answer is 2