c语言，sizeof关键字的用法

Question

# include<stdio.h>
#include<string.h>
#pragma warning(disable:4996)
int main()
{
    int a[] = { 5,4,4,4};
    int k;
    int m, n;
    int sort(int a[]);
    printf("长度%d\n", sizeof(a));
    sort(a);
    return 0;
}
int  sort(int a[]) {
    printf("长度%d\n", sizeof(a) );
    for (int i = 0; i < sizeof(a)/sizeof(int); i++) {
        printf("%d\n", a[i]);
    }
    return 0;
}

为啥两次输出同一个数组，长度变化了

Answer 1

#include <stdio.h>
#include <string.h>
#pragma warning(disable : 4996)
int main()
{
    int a[] = {5, 4, 4, 4};
    int k;
    int m, n;
    int sort(int a[]);
    printf("长度%d\n", sizeof(a)); // 这里的a是数组，所以sizeof(a)返回的是数组的大小
    sort(a);
    return 0;
}
int sort(int a[]) // 当数组当作参数传入函数时，这个参数退化为指针
{
    printf("长度%d\n", sizeof(a)); // 所以这里的sizeof(a)返回的是指针的大小
    for (int i = 0; i < sizeof(a) / sizeof(int); i++)
    {
        printf("%d\n", a[i]);
    }
    return 0;
}

From https://en.cppreference.com/w/c/language/array#Array-to-pointer-conversion

When an array type is used in a function parameter list, it is transformed to the corresponding pointer type: int f(int a[2]) and int f(int* a) declare the same function. Since the function's actual parameter type is pointer type, a function call with an array argument performs array-to-pointer conversion; the size of the argument array is not available to the called function and must be passed explicitly: