关于C语言冒泡排序,让下列四个国家按人口从高到低排

问题遇到的现象和发生背景

到这就想不出来了,有没有大神会的。。
main函数需要呼出bubble_sort_countries

问题相关代码,请勿粘贴截图

#include <stdio.h>

#define COUNTRYNO 4

void bubble_sort_countries(int countries[COUNTRYNO],int n){
int i,j,tmp;
for (i = 0; i < n - 1; i++) {
for (j = 0; j < n - i - 1; j++) {
if (countries[j] > countries[j + 1]) {
tmp = countries[j];
countries[j] = countries[j + 1];
countries[j + 1] = tmp;
}
}
}
}

struct Country {
char country_name[50];
int population;
};

int find_most_populated( struct Country * );

int main( void ) {
struct Country countries[COUNTRYNO] = {
{ "United States", 331 },
{ "China", 1439 },
{ "Japan", 126},
{ "Brazil", 212 }
};
bubble_sort_countries(countries[], 5);
for (i = 0; i < 5; i++) {
printf("%d ", countries[i]);
}

我的解答思路和尝试过的方法
我想要达到的结果

List of most populated countries

  1. Country: China, Population: 1439 million
  2. Country: United States, Population: 331 million
  3. Country: Brazil, Population: 212 million
  4. Country: Japan, Population: 126 million

if (countries[j] > countries[j + 1])
改为
if (countries[j].population < countries[j + 1].population)
temp改为struct Country 类型

#include <stdio.h>
#define COUNTRYNO 4
struct Country {
    char country_name[50];
    int population;
};
void bubble_sort_countries(struct Country countries[COUNTRYNO],int n){
    int i,j;
    struct Country tmp;
    for (i = 0; i < n - 1; i++) {
        for (j = 0; j < n - i - 1; j++) {
            if (countries[j].population < countries[j + 1].population) {
                tmp = countries[j];
                countries[j] = countries[j + 1];
                countries[j + 1] = tmp;
            }
        }
    }
}

int main( void ) 
{
    int i=0;
    struct Country countries[COUNTRYNO] = 
    {
        { "United States", 331 },
        { "China", 1439 },
        { "Japan", 126},
        { "Brazil", 212 }
    };
    bubble_sort_countries(countries, COUNTRYNO);
    for (i = 0; i < COUNTRYNO; i++) {
        printf("Country: %s, Population: %d million\n", countries[i].country_name,countries[i].population);

    }

    return 0;
}