char数组初始化程序中的多余元素错误

Question

我一直在尝试执行以下代码。.但是，我一遍又一遍地遇到相同的错误，我不知道为什么！

我的代码：

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <string.h>

int main(void){

    int randomNum;
    char takenWords[4];
    char words[20]={"DOG", "CAT", "ELEPHANT", "CROCODILE", "HIPPOPOTAMUS", "TORTOISE", "TIGER", "FISH", "SEAGULL", "SEAL", "MONKEY", "KANGAROO", "ZEBRA", "GIRAFFE", "RABBIT", "HORSE", "PENGUIN", "BEAR", "SQUIRREL", "HAMSTER"};


    srand(time(NULL));

    for(int i=0; i<4; i++){
        do{
            randomNum = (rand()%20);
        takenWords[i]=words[randomNum];
        }while((strcmp(&words[randomNum], takenWords) == 0)&&((strcmp(&words[randomNum], &takenWords[i])==0)));
        printf("%s\n", &words[randomNum]);
    }
    getchar();
    return 0;
}

我已经计算了我在数组中输入的元素数量，并且它们不超过20个！

另外，为什么我总是收到“隐式转换丢失整数精度错误”的信息？

Answer 1

我猜您想创建一个指针数组而不是字符数组。

尝试这个：

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <string.h>

int main(void){

    int randomNum;
    const char *takenWords[4];
    const char *words[20]={"DOG", "CAT", "ELEPHANT", "CROCODILE", "HIPPOPOTAMUS", "TORTOISE", "TIGER", "FISH", "SEAGULL", "SEAL", "MONKEY", "KANGAROO", "ZEBRA", "GIRAFFE", "RABBIT", "HORSE", "PENGUIN", "BEAR", "SQUIRREL", "HAMSTER"};


    srand(time(NULL));

    for(int i=0; i<4; i++){
        int dupe=0;
        do{
            randomNum = (rand()%20);
            takenWords[i]=words[randomNum];
            dupe=0;
            for(int j=0;j<i;j++){
                if(strcmp(words[randomNum],takenWords[j])==0)dupe=1;
            }
        }while(dupe);
        printf("%s\n", words[randomNum]);
    }
    getchar();
    return 0;
}

Answer 2

char words[20]={"DOG", "CAT", "ELEPHANT", "CROCODILE", "HIPPOPOTAMUS", "TORTOISE", "TIGER", "FISH", "SEAGULL", "SEAL", "MONKEY", "KANGAROO", "ZEBRA", "GIRAFFE", "RABBIT", "HORSE", "PENGUIN", "BEAR", "SQUIRREL", "HAMSTER"};

仔细查看您声明的数组。它包含什么？

字符串文字（即"DOG"和其他字符串文字）。

看看数组本身，它被声明为存储 char

char words[20]

这就是错误。

要存储字符串文字，您需要pointer to char，即char *

由于您有一个字符串文字数组，因此您需要一个数组 char *

char* words[20]={"DOG", "CAT", "ELEPHANT", "CROCODILE", "HIPPOPOTAMUS", "TORTOISE", "TIGER", "FISH", "SEAGULL", "SEAL", "MONKEY", "KANGAROO", "ZEBRA", "GIRAFFE", "RABBIT", "HORSE", "PENGUIN", "BEAR", "SQUIRREL", "HAMSTER"};

由于您还使用另一个数组来指向字符串文字，因此请以相同的方式声明它们

char* takenWords[4];