array
int a[10]The a in is the array name, which points to the first element in the array. But when it is used as sizeofan operand, or when using an &address, it should abe considered as an array.
Array name and array name take address
Let's look at an example:
int a[10] = {0};
printf("%p, %p\n", a, &a);The print result is 0xbfc077b4, 0xbfc077b4. Both values are the same, but their types are different. aRepresents the address of the first element of the array, type is int*, &arepresents athe address of the array, type is int (*)[10], a pointer to an array containing 10 int elements.
We can use the following example to verify the above conclusion:
int b[10] = {0};
printf("%p, %p, %p, %p\n", b, &b, b+1, &b+1); The print result is 0xbf890214, 0xbf890214, 0xbf890218, 0xbf89023c. a+1The stride is the size of an array element, while &a+1the stride is the size of the entire array.
array as function parameter
When an array is a function parameter, its type degenerates to a pointer. Take a look at an example:
int test_func(int a[])
{
printf("the value is %d\n", sizeof(a));
return 0;
}
int main(int argc, char* argv[])
{
int arr[10] = {0};
test_func(arr);
return 0;
}In the above example, it sizeof(a)is 4, because when the array is used as a function parameter, it degenerates into a pointer, so sizeof(a)it is actually sizeof(int*).