Question

我有这个项目。我有得分，match1，match2的数组，我有球员的名字和他们的国家缩写。那么我有这些缩写和国家的长名称。

我可以将玩家名称和国家缩写放在一起，我也可以将国家缩写和国家长名称放在一起。

但是我不知道如何把国家长名和球员名字放在一个数组中。

所以，我创建了数组（char name_and_country [PLAYERS] [LENGTH_NAME]），我希望这个数组是名称和国家;所以，如果我打印说：

char name_and_country [PLAYERS] [LENGTH_NAME] = {“David Beckham England”，“Wayne Rooney England”等等。

任何人都可以帮助我吗？提前谢谢！

#include <stdlib.h>
#include <stdio.h>
#include <stddef.h>
#define PLAYERS 5
#define NUM_COUNTRIES 3
#define LENGTH_NAME 40
#define LENGTH_CODE 4
#define LENGTH_COUNTRY 20


int main (void)

{   int match1[PLAYERS] = { 0,1,3,2,4};
    int match2[PLAYERS] = { 0,4,0,0,1};
    int goals[PLAYERS] ;

    char name[PLAYERS][LENGTH_NAME] ={"David Beckham","Wayne Rooney","Pirlo", "Del Piero","Lionel Messi"};
    char country_abbreviations[PLAYERS][LENGTH_CODE] = {"ENG","ENG","ITA","ITA","ARG"};
    char country_code[NUM_COUNTRIES][LENGTH_CODE] = {"ARG","ENG","ITA"};
    char country_name[NUM_COUNTRIES][LENGTH_COUNTRY] = {"Argentina", "England","Italy"};
    char name_and_country_code[PLAYERS][LENGTH_NAME];
    char country_code_and_country_name[NUM_COUNTRIES][LENGTH_COUNTRY];
    char name_and_country[PLAYERS][LENGTH_NAME];
    int i, first =1, second= 2;

 for(i=0; i < PLAYERS; i++)
    {
        strcpy (name_and_country_code[i], name[i]);
        strcat (name_and_country_code[i], "  " );
        strcat (name_and_country_code[i], country_abbreviations[i]);
        goals[i]= match1[i] + match2[i];
        printf("Player %s----- score %d:\n", name_and_country_code[i], goals[i]);
    }

Answer 1

使用此代码。我已将代码添加到您的代码中，结果如下。我跑了，它工作得很好..

#include <stdlib.h>
#include <stdio.h>
#include <stddef.h>
#include<string.h>
#define PLAYERS 5

#define NUM_COUNTRIES 3
#define LENGTH_NAME 40
#define LENGTH_CODE 4
#define LENGTH_COUNTRY 20


int main (void)

{   int match1[PLAYERS] = { 0,1,3,2,4};
        int match2[PLAYERS] = { 0,4,0,0,1};
        int goals[PLAYERS] ;

        char name[PLAYERS][LENGTH_NAME] ={"David Beckham","Wayne Rooney","Pirlo", "Del Piero","Lionel Messi"};
        char country_abbreviations[PLAYERS][LENGTH_CODE] = {"ENG","ENG","ITA","ITA","ARG"};
        char country_code[NUM_COUNTRIES][LENGTH_CODE] = {"ARG","ENG","ITA"};
        char country_name[NUM_COUNTRIES][LENGTH_COUNTRY] = {"Argentina", "England","Italy"};
        char name_and_country_code[PLAYERS][LENGTH_NAME];
        char country_code_and_country_name[NUM_COUNTRIES][LENGTH_COUNTRY];
        char name_and_country[PLAYERS][LENGTH_NAME];
        int i, first =1, second= 2;

        for(i=0; i < PLAYERS; i++)
        {
                strcpy (name_and_country_code[i], name[i]);
                strcat (name_and_country_code[i], "  " );
                strcat (name_and_country_code[i], country_abbreviations[i]);
                goals[i]= match1[i] + match2[i];
                printf("Player %s----- score %d:\n", name_and_country_code[i], goals[i]);
        }
        for(i=0; i < PLAYERS; i++)
        {
                strcpy (name_and_country[i], name[i]);
                strcat (name_and_country[i], "  " );
                char country[LENGTH_COUNTRY];
                strcpy(country,"DEFAULT COUNTRY"); // Used when player has a invalid country code
                int j;
                for(j=0;j<NUM_COUNTRIES;j++)
                {
                        if(strcmp(country_abbreviations[i],country_code[j])==0)
                                strcpy(country,country_name[j]);
                }
                strcat(name_and_country[i],country);
                printf("%s\n",name_and_country[i]);
        }
}

如何在C编程中组合两个字符串数组

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