char数组初始值设定项错误中的多余元素

Question

我一直在尝试执行以下代码..但是，我一遍又一遍地得到同样的错误，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，它们不超过20！

另外，为什么我一直得到'隐式转换失去整数精度错误'？

Answer 1

我想你想要制作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"};

查看您已仔细声明的数组。 它包含什么？

字符串文字（即"DOG"和其他字符串文字）。

查看数组本身，声明它存储char

char words[20]

那是错误。

---

要存储字符串文字，需要pointer to char ，即char *

由于你有一个字符串文字数组，你需要一个char *数组

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指向字符串文字，因此以相同的方式声明它们

char* takenWords[4];

Answer 3

在你的代码中， char words[20]是一个数组，用于保存20个字符但不包含20个不同字符串的单个字符串/单词。

然而，为了声明20个不同的字符串，使用2D数组，这样char字[20] [20]然后继续声明你选择的字符串/单词最多20个长度。

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

注意： char words[20][20]表示一个数组，用于容纳20个最大长度为20个字符的字符串......

通常char words[m][n]其中m，n是整数，包含m个最大长度的字符串（n-1个字符）+（终止空字符） 。

Answer 4

char arr [NUMBER_OF_STRING] [MAX_STRING_SIZE]声明它的最大字符串大小